The question is what do you mean by a "model of ZFC". If you mean the axioms as we enumerate them, in the meta-theory, this may or may not be the same thing as those of the universe of sets in which we are looking. The same can be told about the inference rules of FOL.
But since we can code everything into integers, that means that if we have a model whose integers are standard (read: agree with the meta-theory), then these problems go away.
What does that mean? Well, if the model was a model of ZFC, then it proves the completeness theorem, and since the arithmetic statement "ZFC is consistent" is true, that means that we can find a model of ZFC, and that it is the same ZFC as our meta-theory's one.
You might want to argue that the relation of that model might not be "correct" in some way, but we can make this model internally countable and the relation is then coded by a subset of $\omega$. So it's all good.
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
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.