= Solution
Let $G=\operatorname{Gal}(K/k)$. 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 that
$$
a\equiv0\pmod{\mathfrak P}\quad(\mathfrak P\in\mathcal C),\qquad
a\equiv1\pmod{\mathfrak Q}\quad(\mathfrak Q\notin\mathcal C,\ \mathfrak Q\mid\mathfrak p).
$$
The <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 \b[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.
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