Existence in ZF of a set with countable power set Is it consistent with ZF for there to exist a set $S$ such that the power set $P(S)$ is countable? If so, what is the weakest form of the axiom of choice needed to prove that no such set exists?

Asaf has answered the question. Let me just add that even more is true: Provably in $\mathsf{ZF}$, If $A$ is a set, and $\aleph_0\le|\mathcal P(A)|$, then in fact $|\mathbb R|=2^{\aleph_0}\le|\mathcal P(A)|$. This is a result of Kuratowski. What one proves is that from an injection from $\mathbb N$ into $\mathcal P(A)$, an infinite sequence of pairwise disjoint subsets of $A$ can be obtained (this takes some work), so there is a surjection from $A$ onto $\mathbb N$, and from this we easily get an injection of $\mathcal P(\mathbb N)$ into $\mathcal P(A)$.
It may well be the case that we have an infinite set $A$ such that $\aleph_0\not\le|\mathcal P(A)|$, though we always have $\aleph_0\le|\mathcal P(\mathcal P(A))|$ for $A$ infinite. 
