We'll go with your first definition. Note that your two definitions are not equivalent; for example $P(6,3)$ produces the probabilities $\tfrac25,\tfrac3{10},\tfrac15,\tfrac1{10}$ for the first definition but $\tfrac13,\tfrac49,\tfrac19,\tfrac19$ for the other.
Let's first find how many $1$'s there are in $P(n,k)$. First note, with
stars and bars we can deduce that $$|P(n,k)|={n-1\choose k-1}$$
Now let me introduce you to my notation: I denote
$$(i,P(n,k)):=\{(i,j_1,j_2,\cdots,j_k)|(j_1,\cdots,j_k)\in P(n,k)\}$$
So, for example, $P(3,2)=\{(1,2),(2,1)\}$ and thus $(4,P(3,2))=\{(4,1,2),(4,2,1)\}$. Now notice the following:
$$P(n,k)=\bigcup_{i=1}^{n-k+1} (i,P(n-i,k-1))\tag{$*$}$$
To illustrate this fact, take your $P(5,3)$ example and see;
\begin{align}
P(5,3)=&\{(1,1,3),(1,2,2),(1,3,1)\}\tag{that's $(1,P(4,2))$}\\
&\cup\{(2,1,2),(2,2,1)\}\tag{that's $(2,P(3,2))$}\\
&\cup\{(3,1,1)\}\tag{that's $(3,P(2,2))$}
\end{align}
Try this yourself on paper and you'll see why expression $(*)$ is trivial.
Now let us introduce another notation: let's write $P_a(n,k)$ for the number of $a$'s that occur in $P(n,k)$. So for example, $P_1(5,3)=9$. Now with formula $(*)$, we see (for $k>1$):
\begin{align}
P_a(n,k)&=|P(n-a,k-1)|+\sum_{i=1}^{n-k+1}P_a(n-i,k-1)\\
&={n-a-1\choose k-2}+\sum_{i=1}^{n-k+1}P_a(n-i,k-1)
\end{align}
This formula is useful, because we can do induction with it to prove possible conjectures we have. For example, calculating some values for $P_1(n,k)$ leads quickly to a pattern;
For all $n,k$ with $n\neq 1$, we have $P_1(n,k)=k\cdot{n-2\choose k-2}$
Proof. First note that this theorem uses the fact that ${a\choose b}=0$ if $b<0$ or $b>a$. We will now use induction to $n$ to prove our statement. First, if $n=2$, then the expression says $P_1(2,k)=k\cdot {0\choose k-2}$. This is easy to confirm numerically: $P(2,1)=\{(2)\}$ which has $P_1(2,1)=1\cdot {0\choose -1}=0$ ones in it; and $P(2,2)=\{(1,1)\}$ has $P_1(2,2)=2\cdot {0\choose 0}=2$ ones in it. All good so far! Now assume the theorem is true for $(1,k),(2,k),\cdots,(n-1,k)$. Then we see, for $k\geq 3$:
\begin{align}
P_1(n,k)&={n-2\choose k-2}+\sum_{i=1}^{n-k+1}P_1(n-i,k-1)\\
&={n-2\choose k-2}+\sum_{i=1}^{n-k+1}(k-1){n-i-2\choose k-3}\\
&={n-2\choose k-2}+(k-1)\sum_{i=1}^{n-k+1}{n-i-2\choose k-3}\\
&={n-2\choose k-2}+(k-1)\sum_{i=k-3}^{n-3}{n-(i-k+4)-2\choose k-3}\\
&={n-2\choose k-2}+(k-1)\sum_{i=k-3}^{n-3}{(n-3)-i+(k-3)\choose k-3}\\
&={n-2\choose k-2}+(k-1)\sum_{i=k-3}^{n-3}{i\choose k-3}\\
&={n-2\choose k-2}+(k-1){n-2\choose k-2}\\
&=k\cdot {n-2\choose k-2}
\end{align}
Where we've used the Hockey-stick identity in the second-last line. For $k=1$, the expression is obvious; $P_1(n,1)=0$, since $P(n,1)=\{(n)\}$ and $n\neq 1$. Now we have $k=2$ left; but this case is trivial too, since $P(n,2)=\{(1,n-1),(2,n-2),\cdots,(n-2,2),(n-1,1)\}$, so $P_1(n,2)=2$. Thus, by the principle of mathematical induction, we've proved the statement.
With this statement, we already have a major result: the probability of having a $1$ in $P(n,k)$ is $$\frac{k{n-2\choose k-2}}{k{n-1\choose k-1}}=\frac{k-1}{n-1}$$
We divide by $k{n-1\choose k-1}$ because $|P(n,k)|={n-1\choose k-1}$ and each partition contains $k$ numbers.
We can actually do the
exact same thing for other numbers.
For all $n,k,a$ with $n>a$, we have $P_a(n,k)=k\cdot{n-a-1\choose k-2}$
First note that $P_a(n,1)=0$ (except when $n=a$, then $P_a(n,1)=1$). This is true because $P(n,1)=\{(n)\}$ (hence the exception). Also, $P_a(n,n-i)=0$ if $i<a-1$, since if we want to partition $n$ in sums of $n-i$ terms, and we want to use one term $a$, then we have to partition $n-a$ in $n-i-1$ terms; this will obviously not work (because $n-a<n-i-1$ so there should be at least one term less than $1$, impossible). Now we can do induction again (a little quicker this time):
\begin{align}
P_a(n,k)&={n-a-1\choose k-2}+\sum_{i=1}^{n-k+1}P_a(n-i,k-1)\\
&={n-a-1\choose k-2}+(k-1)\sum_{i=1}^{n-k+1}{n-i-a-1\choose k-3}\\
&={n-a-1\choose k-2}+(k-1)\sum_{i=k-a-2}^{n-a-2}{i\choose k-3}\\
&={n-a-1\choose k-2}+(k-1)\sum_{i=k-3}^{n-a-2}{i\choose k-3}\\
&={n-a-1\choose k-2}+(k-1){n-a-1\choose k-2}\\
&=k{n-a-1\choose k-2}
\end{align}
You shouldn't do induction this loosely normally; I do it here to reduce the length of the post. Try to convince yourself of this theorem by doing the proof rigorously.
And so we arrive at our final result:
The probability of having an $a$ in $P(n,k)$ is exactly
$$\frac{k{n-a-1\choose k-2}}{k{n-1\choose k-1}}=\frac{{n-a-1\choose k-2}}{{n-1\choose k-1}}$$
We don't simplify this further with factorials, since ${a\choose b}=\frac{a!}{b!(a-b)!}$ is only true for $0\leq b\leq a$.
So in case of $P(5,3)$, we see the chance of having a $3$ is
$$\frac{{5-3-1\choose 3-2}}{{5-1\choose 3-1}}=\frac{{1\choose 1}}{{4\choose 2}}=\frac16$$
