Kunen is doing that unfortunate thing, where you define cardinals only for well-ordered sets without choice.
And I say unfortunate, since it causes unnecessary confusion. While I agree that in order to define cardinals without choice we need to talk about the von Neumann hierarchy first, and this might not be the best starting point for a student, it is also possible to remark that we will address this definition in the future. But most people, when thinking about the foundation of set theory, are so immersed in the "lack of importance of non-AC to the discussion", that by the time they reached this point of the von Neumann hierarchy, the axiom of choice is deeply ingrained into their text and they forgot all about those "silly, non-well orderable sets".
Moving on. Yes. According to the approach taken by Kunen, the axiom of choice is needed to prove that $|\mathcal P(\omega)|$ is an uncountable cardinal, because cardinal means necessarily well-orderable.
But choice is only needed to show that $|\mathcal P(\omega)|$ is even a thing, not that it is uncountable. The fact is that $\sf ZF$ proves quite directly Cantor's theorem, so $\omega\prec\mathcal P(\omega)$, which means that the latter is most definitely uncountable. It just might not have a cardinal, when going with Kunen's approach here.
So what about uncountable cardinals in general? Well, the [well-ordered] cardinals are generated by means of Hartogs' theorem, unions, and transfinite recursion. They are always ordinals. They only exits, as follows from the axioms of $\sf ZF$.
So we get two ways of generating larger and larger sets: Hartogs' operation, and the power set operation. Even assuming choice these may give us two different functions ($\aleph$ and $\beth$ numbers).