 Solution

Let $G=\mathrm{Gal}(K/k)$. Each field automorphism preserves the ring of integers and the base prime ideal, so it permutes the prime ideals above $\mathfrak{p}$. Suppose there were more than one orbit, and fix one orbit $\mathcal{C}$. By the Chinese remainder theorem for ideals, choose $a\in {\mathcal{O}}_K$ such thatThe field norm $N_{K/k}(a)={\prod }_{\sigma \in G}\sigma (a)$ lies in ${\mathcal{O}}_k$. It lies in every prime ideal of $\mathcal{C}$, since for a fixed member each conjugate of $a$ is zero modulo that member. Thus its contraction lies in $\mathfrak{p}$. On the other hand, for $\mathfrak{Q}$ outside $\mathcal{C}$, every ${\sigma }^{-1}(\mathfrak{Q})$ is also outside it, so every $\sigma (a)$ equals one modulo $\mathfrak{Q}$. The product is consequently one modulo $\mathfrak{Q}$. This contradicts its membership in $\mathfrak{p}{\mathcal{O}}_K\subseteq \mathfrak{Q}$. Therefore the Galois action on the primes above $\mathfrak{p}$ is transitive. The normal extension of number fields is separable, so its conjugates are exactly the elements of $G$ used in the product.