Weak Power Hypothesis and Dependent Choice

Consider in $$\newcommand{\ZF}{{\sf (ZF)}}\ZF$$ the following statement:

Weak Power Hypothesis (WPH): if $$X,Y$$ are sets and there is a bijection between $$\newcommand{\P}{{\cal P}}\P(X)$$ and $$\P(Y)$$, then there is a bijection between $$X$$ and $$Y$$.

The Axiom of Dependent Choice states that

(DC): if $$X$$ is a set and $$R\subseteq X\times X$$ such that for all $$a\in X$$ there is $$b\in X$$ such that $$(a,b)\in R$$, then there is a function $$s:\omega\to X$$ such that $$\big(s(n),s(n+1)\big) \in R$$ for all $$n\in \omega$$.

Does (WPH) imply (DC)?

Note: It appears to be open whether (WPH) implies the much stronger Axiom of Choice (AC).

I've reduced the problem to "every infinite set is Dedekind infinite", which is a consequence of WPH if my memory serves me right (this was answered before on the site).

We'll show that well-ordered choice holds, which is known to imply Dependent Choice. In fact, we will argue the equivalent, $$\forall A(\aleph(A) = \aleph^*(A))$$.

Suppose that $$A$$ is an infinite set, then the following holds: $$\aleph(A) = \aleph(A\times\omega),$$ since $$A$$ is Dedekind infinite. And of course, $$|A\times\omega|=|A\times\omega|\cdot 2.$$

If $$\kappa<\aleph^*(A\times\omega)$$, then:

$$2^{A\times\omega}\leq 2^{A\times\omega}\cdot 2^\kappa = 2^{A\times\omega + \kappa}\leq 2^{A\times\omega}\cdot 2^{A\times\omega} = 2^{A\times\omega}.$$

By WPH, $$|A\cdot\omega|=|A\cdot\omega+\kappa|$$, so in particular, $$\kappa<\aleph(A\cdot\omega)=\aleph(A)$$.

Therefore, $$\aleph(A) = \aleph^*(A\cdot\omega)$$, and so we get $$\aleph(A) =\aleph^*(A)$$, so Dependent Choice holds.