Without AC how to construct a surjection between powerset of a cardinal to another cardinal? With Axiom of Choice, we could have some consequences between $2^{\aleph_{\alpha}}$ and $\aleph_{\alpha}$ or some other cardinals without having to construct a map. If without AC, the cardinality of $2^{\aleph_{\alpha}}$ is not defined. Can we construct a surjection from $\mathcal{P} (\aleph_{\alpha})$ to $\aleph_{\alpha+1}$?
 A: First, note that your claim

without AC, the cardinality of $2^{\aleph_\alpha}$ is not defined

is false. The cardinality of a set is definable, with or without choice (although without choice, the way we define "cardinality" is more complicated). What is true is that without choice, we can't prove that the cardinality of $2^\kappa$ is an $\aleph$-number.
(Incidentally, even with choice we can't prove which $\aleph$-number something like $2^{\aleph_0}$ is - e.g. it is consistent with ZFC that $2^{\aleph_0}=\aleph_1$, but it's also consistent with ZFC that $2^{\aleph_0}=\aleph_{\omega^2+17}$.)

Now on to your actual question:
Yes, ZF alone proves that there is a surjection from $\mathcal{P}(\aleph_\alpha)$ to $\aleph_{\alpha+1}$.
The key is the following observation: that $\aleph_\alpha$ is infinite well-orderable by definition, and $\aleph_{\alpha+1}$ is the cardinality of the set of isomorphism types of well-orderings of subsets of $\aleph_\alpha$. To prove this, first show that $\aleph_{\alpha+1}$ is the cardinality of the set of ordinals of cardinality $\le\aleph_\alpha$ (think about why $\aleph_1$ is the cardinality of the set of countable ordinals), and then try to show that every ordinal of cardinality $\le\kappa$ can be represented by a well-ordering of a subset of $\kappa$.
Given the claim above, we now build the desired surjection $o$. Since $\aleph_\alpha$ is infinite and well-orderable, there is a bijection $b$ from $\mathcal{P}(\aleph_\alpha)$ to $\mathcal{P}(\aleph_\alpha\times\aleph_\alpha)$ (exercise). For $A\in\aleph_\alpha$, let $o(A)$ be defined as follows:


*

*If $b(A)$ is a well-ordering of a subset of $\aleph_\alpha$, then $o(A)$ is the ordertype of $b(A)$.

*Otherwise, $o(A)$ is the ordertype of the usual ordering on $\aleph_\alpha$.
The map $o$ is (exercise) a surjection from $\mathcal{P}(\aleph_\alpha)$ to the set of well-orderings of subsets of $\aleph_\alpha$; so we're done.

Interestingly, ZF does not prove that there is an injection from $\aleph_{\alpha+1}$ to $\mathcal{P}(\aleph_\alpha)$! E.g. the axiom of determinacy implies that there is no injection from $\aleph_1$ to $2^{\aleph_0}$.
