If power sets $\mathcal P(A)$ and $\mathcal P(B)$ have the same cardinality, does that imply that sets $A$ and $B$ have the same cardinality?
The converse is obviously true. Namely, if $|A| = |B|$, then there is a bijection $f: A \to B$, from which another bijection $F: \mathcal P(A) \to \mathcal P(B)$ can be trivially constructed, such that
$$b \in F(X) \iff f^{-1}(b) \in X$$ or equivalently $$a \in F^{-1}(Y) \iff f(a) \in Y,$$
where $a\in A$, $b\in B$, $X \in \mathcal P(A)$ (i.e. $X \subseteq A$) and $Y \in \mathcal P(B)$ (i.e. $Y \subseteq B$).
However, this seems to offer no insight whether $|\mathcal P(A)|=|\mathcal P(B)|$ implies $|A|=|B|$. Is it possible for two sets to have different cardinalities, but such that the cardinalities of their respective power sets are the same?
It is obviously not possible for finite sets. However, for infinite sets, the situation is not so clear. If the generalized continuum hypothesis (which is independent from ZFC) is added as an axiom, then it is also not possible because then $2^{\aleph_\alpha} = \aleph_{\alpha+1}$, so if $|A|=\aleph_\alpha$ and $|B|=\aleph_\beta$, then simply $|\mathcal P(A)| = \aleph_{\alpha+1}$ and $|\mathcal P(B)| = \aleph_{\beta+1}$, and therefore: $$|A|\ne|B| \implies \aleph_\alpha \ne \aleph_\beta \implies \alpha \ne \beta \implies \alpha + 1 \ne \beta + 1 \implies \aleph_{\alpha+1} \ne \aleph_{\beta+1} \implies |\mathcal P(A)| \ne |\mathcal P(B)|$$
However, it is not clear whether it is possible in the case of negation of the generalized continuum hypothesis.
