No (assuming ZFC is consistent). This is essentially immediate from the compactness theorem. For instance, add uncountably many constant symbols $c_i$ to the language of ZFC, and add axioms saying that $c_i\neq c_j$ for each $i\neq j$ and $c_i\in\omega$ for each $i$. Every finite subset of the resulting theory is consistent, since it will only involve finitely many of the $c_i$ and so we can send those $c_i$ to distinct natural numbers in some model of ZFC and the rest to $0$. So by compactness, this theory has a model $M$. Then $M$ is a model of ZFC such that $\omega^M$ has (externally) uncountably many elements.
Yes, $\omega^M$ satisfies PA. More generally, suppose $X$ is any first-order structure internal to $M$. Then a simple induction on formulas shows that for any first-order formula $\varphi(x_1,\dots,x_n)$ and any $a_1,\dots,a_n\in X$, $M\models(X\models\varphi(a_1,\dots,a_n))$ iff $X\models\varphi(a_1,\dots,a_n)$. (To be clear, when I say $X\models\varphi(a_1,\dots,a_n)$, I really mean $X'\models\varphi(a_1,\dots,a_n)$ where $X'=\{a\in M:M\models a\in X\}$ equipped with the first-order structure defined using the internal first-order structure of $X'$.)
However, $\omega^M$ cannot be an arbitrary nonstandard model of PA. For instance, since ZFC proves the consistency of PA, $M\models(\omega\models Con(PA))$ and thus $\omega^M$ must satisfy $Con(PA)$.
(There is an important subtlety here, which is that the argument of the first paragraph only applies when $\varphi$ is an actual first-order formula, i.e. an external one in the real universe, rather than a first-order formula internal to $M$. If $\omega^M$ is nonstandard, then $M$ will have "first-order formulas" whose length is nonstandard and therefore are not actually formulas from the external perspective. This means, for instance, that if $X$ is a structure internal to $M$ which is externally a model of PA, then $M$ may not think $X$ is a model of PA, since $M$ has nonstandard axioms of PA which $X$ may not satisfy. See this neat paper for some dramatic ways that things like this can occur.)
Best Answer
Essential point is that $\mathcal{P}(X)$ doesn’t necessarily include “every subset of $X$ you can imagine”. Rather, its members are just the subsets of $X$ that actually exist in the model. So, in a model with non-standard $\omega$, there is no set whose members are only the initial segment of $\omega$ that’s order-isomorphic to “standard $\mathbb{N}$”