I remember when you last asked this question, I thought there was a counterexample and wrote it down. It didn’t end up working, and I couldn’t figure out a solution in the end. Let me try my hand again this time, instead with a proof that the result is true.
Uniqueness: Any $\overline{\pi}: M(A) \to M(B)$ extending $\pi$ must satisfy $\overline{\pi}(m)\pi(a)b = \pi(ma)b$. Since $\pi(A)B$ is dense in $B$, and $B$ is an essential ideal in $M(B)$, this shows $\overline{\pi}$ is the same for any possible extension $\overline{\pi}$, i.e., the extension $\overline{\pi}$ is unique.
Existence: For this, we need a more concrete description of $M(A)$. Recall that a double centralizer of $A$ is a pair of bounded maps $(L, R)$ from $A$ to itself, s.t. $L(ab) = L(a)b$, $R(ab) = aR(b)$, and $aL(b) = R(a)b$ for all $a, b \in A$. Recall that if $(L, R)$ is a double centralizer, then $\|L\| = \|R\|$ and we let the norm of $(L, R)$ be $\|L\|$, or equivalently $\|R\|$. Recall that $M(A)$ is the linear space of all double centralizers. It is a $C^\ast$-algebra under the norm above, the multiplication $(L, R)(L’, R’) = (LL’, R’R)$, and the involution $(L, R)^\ast = (R^\ast, L^\ast)$, where for a linear map $T: A \to A$, we let $T^\ast(a) = T(a^\ast)^\ast$. For each $a \in A$, let $L_a(b) = ab$ and $R_a(b) = ba$. Then $(L_a, R_a)$ is a double centralizer and $\iota: A \to M(A)$ given by $\iota(a) = (L_a, R_a)$ is an inclusion of $C^\ast$-algebra, whose range we shall identify with $A$ and is an essential ideal of $M(A)$. We note that if $m = (L, R) \in M(A)$ and $a \in A$, then $ma = L(a)$ and $am = R(a)$.
Now back to the problem at hand. For each $m = (L, R) \in M(A)$, we define $L_m: \pi(A)B \to B$ by,
$$L_m(\sum_{i=1}^n \pi(a_i)b_i) = \sum_{i=1}^n \pi(L(a_i))b_i$$
We need to first show that this is well-defined, that is, if $\sum_{i=1}^n \pi(a_i)b_i = \sum_{j=1}^m \pi(a’_j)b’_j$, then $\sum_{i=1}^n \pi(L(a_i))b_i = \sum_{j=1}^m \pi(L(a’_j))b’_j$. Let $\{u_\lambda\}$ be an approximate identity of $A$, then,
$$\begin{split}
\sum_{i=1}^n \pi(L(a_i))b_i &= \lim_\lambda \sum_{i=1}^n \pi(L(u_\lambda a_i))b_i\\
&= \lim_\lambda \pi(L(u_\lambda))\sum_{i=1}^n \pi(a_i)b_i\\
&= \lim_\lambda \pi(L(u_\lambda))\sum_{j=1}^m \pi(a’_j)b’_j\\
&= \lim_\lambda \sum_{j=1}^m \pi(L(u_\lambda a’_j))b’_j\\
&= \sum_{j=1}^m \pi(L(a’_j))b’_j
\end{split}$$
We now show that $L_m$ is bounded. Indeed,
$$\begin{split}
\|\sum_{i=1}^n \pi(L(a_i))b_i\|^2 &= \|\begin{pmatrix} \pi(ma_1) & \cdots & \pi(ma_n) \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}\|^2\\
&= \|\begin{pmatrix} b_1^\ast & \cdots & b_n^\ast \end{pmatrix} \begin{pmatrix} \pi(a_1^\ast m^\ast) \\ \vdots \\ \pi(a_n^\ast m^\ast) \end{pmatrix} \begin{pmatrix} \pi(ma_1) & \cdots & \pi(ma_n) \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}\|\\
&= \lim_\lambda \|\begin{pmatrix} b_1^\ast & \cdots & b_n^\ast \end{pmatrix} \begin{pmatrix} \pi(a_1^\ast u_\lambda m^\ast) \\ \vdots \\ \pi(a_n^\ast u_\lambda m^\ast) \end{pmatrix} \begin{pmatrix} \pi(m u_\lambda a_1) & \cdots & \pi(m u_\lambda a_n) \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}\|\\
&= \lim_\lambda \|\begin{pmatrix} b_1^\ast & \cdots & b_n^\ast \end{pmatrix} \begin{pmatrix} \pi(a_1^\ast) \\ \vdots \\ \pi(a_n^\ast) \end{pmatrix} \pi(u_\lambda m^\ast) \pi(m u_\lambda) \begin{pmatrix} \pi(a_1) & \cdots & \pi(a_n) \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}\|\\
&= \lim_\lambda \|\begin{pmatrix} b_1^\ast & \cdots & b_n^\ast \end{pmatrix} \begin{pmatrix} \pi(a_1^\ast) \\ \vdots \\ \pi(a_n^\ast) \end{pmatrix} \pi(u_\lambda m^\ast m u_\lambda) \begin{pmatrix} \pi(a_1) & \cdots & \pi(a_n) \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}\|\\
&\leq \lim_\lambda \|m\|^2 \|\begin{pmatrix} b_1^\ast & \cdots & b_n^\ast \end{pmatrix} \begin{pmatrix} \pi(a_1^\ast) \\ \vdots \\ \pi(a_n^\ast) \end{pmatrix} \pi(u_\lambda^2) \begin{pmatrix} \pi(a_1) & \cdots & \pi(a_n) \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}\|\\
&= \lim_\lambda \|m\|^2\|\begin{pmatrix} b_1^\ast & \cdots & b_n^\ast \end{pmatrix} \begin{pmatrix} \pi(a_1^\ast u_\lambda) \\ \vdots \\ \pi(a_n^\ast u_\lambda) \end{pmatrix} \begin{pmatrix} \pi(u_\lambda a_1) & \cdots & \pi(u_\lambda a_n) \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}\|\\
&= \|m\|^2\|\begin{pmatrix} b_1^\ast & \cdots & b_n^\ast \end{pmatrix} \begin{pmatrix} \pi(a_1^\ast) \\ \vdots \\ \pi(a_n^\ast) \end{pmatrix} \begin{pmatrix} \pi(a_1) & \cdots & \pi(a_n) \end{pmatrix} \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}\|\\
&= \|m\|^2\|\sum_{i=1}^n \pi(a_i)b_i\|^2
\end{split}$$
Thus, $\|L_m\| \leq \|m\|$. As $\pi(A)B$ is dense in $B$, we may extend $L_m$ to a bounded linear map $B \to B$, which we shall still denote by $L_m$.
Note that, as $\pi(A)B$ is dense in $B$, by taking adjoints, we also have $B\pi(A)$ is dense in $B$. Thus, we similarly define $R_m: B\pi(A) \to B$,
$$R_m(\sum_{i=1}^n b_i\pi(a_i)) = \sum_{i=1}^n b_i\pi(R(a_i))$$
By the same argument, $R_m$ is well-defined and $\|R_m\| \leq \|m\|$, so it may be extended to $R_m: B \to B$.
We now claim that $(L_m, R_m)$ is a double centralizer of $B$. That is, we need to verify,
- $L_m(xy) = L_m(x)y$;
- $R_m(xy) = xR_m(y)$;
- $xL_m(y) = R_m(x)y$.
To prove $1$, by linearity and density, the following calculation is sufficient,
$$L_m(\pi(a)bb’) = \pi(L(a))bb’ = L_m(\pi(a)b)b’$$
Similarly, for $2$,
$$R_m(b’b\pi(a)) = bb’\pi(R(a)) = b’R_m(b\pi(a))$$
And for $3$,
$$\begin{split}
b’\pi(a’)L_m(\pi(a)b) &= b’\pi(a’)\pi(L(a))b\\
&= b’\pi(a’L(a))b\\
&= b’\pi(R(a’)a)b\\
&= b’\pi(R(a’))\pi(a)b\\
&= R_m(b’\pi(a’))\pi(a)b
\end{split}$$
Thus, $(L_m, R_m) \in M(B)$. So $\overline{\pi}: M(A) \to M(B)$, $\overline{\pi}(m) = (L_m, R_m)$ is a well-defined contractive linear map. If $m = a \in A$, it is straightforward to verify, from the definitions, that $L_m = L_{\pi(a)}$ and similarly $R_m = R_{\pi(a)}$, so $\overline{\pi}$ extends $\pi$. That $\overline{\pi}$ is multiplicative is easy to verify from the definition. That it preserves the adjoint can be verified by, (recall that $m^\ast = (R^\ast, L^\ast)$),
$$\begin{split}
L_{m^\ast}(\pi(a)b) &= \pi(R^\ast(a))b\\
&= \pi(R(a^\ast)^\ast)b\\
&= [b^\ast \pi(R(a^\ast))]^\ast\\
&= R_m(b^\ast \pi(a^\ast))^\ast\\
&= R_m^\ast(\pi(a)b)
\end{split}$$
Since $\overline{\pi}(m)^\ast = (L_m, R_m)^\ast = (R_m^\ast, L_m^\ast)$, the above shows $\overline{\pi}(m^\ast) = \overline{\pi}(m)^\ast$. Thus, $\overline{\pi}: M(A) \to M(B)$ is a $\ast$-homomorphism extending $\pi$, as desired.
I know last time I was the one who said the two notions of non-degeneracy are different when it comes to $B(H)$, but I should remark that that is only the case for $B = B(H)$, i.e., $\pi: A \to B(H) = M(B(H))$ satisfying $\pi(A)B(H)$ is dense in $B(H)$ is not the same (in fact, strictly stronger) than $\pi(A)H$ being dense in $H$. For example, $A = K(H)$, $\pi: K(H) \to B(H)$ being the natural inclusion satisfies $\pi(A)H = H$ but $\pi(A)B(H) = K(H)$ is certainly not dense in $B(H)$.
However, if $B = K(H)$, the two notions are actually the same, i.e., if $\pi: A \to B(H) = M(K(H))$, then $\pi(A)K(H)$ being dense in $K(H)$ is equivalent to $\pi(A)H$ being dense in $H$. Indeed, if $\overline{\pi(A)K(H)} = K(H)$, then $\overline{\pi(A)H} = \overline{\overline{\pi(A)K(H)}H} = \overline{K(H)H} = H$. For the converse, we note that $\overline{\pi(A)K(H)}$ is a closed right ideal of $K(H)$. By similar proof as the proof that $K(H)$ is simple, one may show that if $I \subset K(H)$ is a closed right ideal, then $I$ consists of all those operators in $K(H)$ whose ranges lie within $\overline{IH}$. Thus, applying this to $I = \overline{\pi(A)K(H)}$, we see that, as $\pi(A)H$ is dense in $H$, $IH \supset \pi(A)H$ is dense in $H$, whence $I$ consists of all compact operators whose ranges lie within $\overline{IH} = H$, so all compact operators, i.e., $\overline{\pi(A)K(H)} = I = K(H)$.
