This is addressing two questions of user18921 asked in the comments to Asaf's answer.
a) The reason why GCH gives long chains in $\mathcal P(X)$ is the following:
For an infinite cardinal $\kappa$ let $L=\{0,1\}^\kappa$ and let $B$ be the set of all sequences in $\{0,1\}^\kappa$ that are eventually 0 (or constant, it doesn't matter). The size of $B$ is $\sup\{2^\lambda:\lambda<\kappa\}$.
$L$ is linearly ordered by the lexicographic ordering and $B$ is dense in $L$. Assuming GCH we have
$2^\lambda\leq\kappa$ for all $\lambda<\kappa$ (this is all of GCH that we need).
Hence $|B|=\kappa$. Let $X=B$. Now the family of sets $\{x\in B:x<y\}$, $y\in L$, is a chain in $\mathcal P(X)$ of size $2^{|X|}$.
b) (2) is equivalent to (2'). One half of this is already hidden in my answer a):
If there is an infinite linear order of size $|X|$ whose completion is of size $2^{|X|}$, then $\mathcal P(X)$ has a chain of length $2^{|X|}$.
On the other hand, assume $\mathcal P(X)$ has a chain $\mathcal C$ of length $2^{|X|}$. We may assume that $\mathcal C$ is closed under infinite intersections
and that it contains $X$.
Now for each $x\in X$ consider the first $A\in\mathcal C$ with $x\in A$
and call it $A_x$. By our assumptions on $\mathcal C$, $A_x$ exists for all $x\in X$. It is just the intersection of all $A\in\mathcal C$ with $x\in A$.
Now $\mathcal A=\{A_x:x\in X\}$ is a dense subset of the linearly ordered set $\mathcal C$ ($\mathcal C$ is ordered by $\subseteq$, of course).
$\mathcal A$ is a linear order of size $|X|$ with a completion of size $2^{|X|}$.
The equivalence of (2) and (2') was observed independently by Baumgartner and Mitchell, if I remember correctly.