Uncountable ordinals without power set axiom

Question

Assume $M$ is a set, in which all axioms of $ZF - P + (V=L)$ hold. Does then $M$ believe that there exists an uncountable ordinal? I mean, why should the class of all countable ordinal numbers be a set?

Answer 1

No. One cannot prove that $\omega_1$ exists in ZF - P, with or without V = L.

The set $HC$ of hereditarily countable sets always satisfies ZF - P. (This is straightforward to check axiom per axiom.) Of course, $\omega_1 \notin HC$ but moreover every set in $HC$ is countable or finite as witnessed by a function in $HC$. Therefore, $HC$ knows that every set in $HC$ is countable and hence $HC$ is a model of ZF - P + "every set is at most countable."

Throwing in V = L into the mix doesn't help. Indeed, $HC^L = L_{\omega_1^L}$ (see my answer to your recent MO question for a proof) is a model of ZF - P + V = L + "every set is at most countable."