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, $$|A\times\omega|=|A\times|\cdot 2|.$$

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.

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.