= Solution
\b[True in ZF.] Let $A$ be a set that is not finite. For each <natural number> $n$, form
$$
E_n=\{B\subseteq A:|B|=n\}.
$$
Each $E_n$ is a subset of $\mathcal P(A)$ and is nonempty: a nonfinite set has an $n$-element subset for every finite $n$, by finite induction. The sets $E_n$ are pairwise disjoint, hence pairwise distinct. Consequently
$$
n\longmapsto E_n
$$
is an <injection> from $\mathbb N$ into $\mathcal P(\mathcal P(A))$. This is a <double-power-set countability bound without choice>.
Part (d), applied to the set $\mathcal P(A)$, says that its <power set> is not countably infinite. Combining that result with the displayed <injection> gives
$$
\boxed{\aleph_0<2^{2^\kappa}.}
$$
There is no hidden <countable> choice: the proof uses the entire definable family of $n$-element subsets, rather than choosing one subset for each $n$.
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