= Solution
Write $G=\operatorname{Gal}(F/K)$ and fix a <prime ideal> $\mathfrak q$ above $\mathfrak p$. Suppose there is another prime $\mathfrak r$ above $\mathfrak p$ outside the $G$-orbit of $\mathfrak q$. Distinct nonzero <prime ideals> in the <ring of integers> are distinct <maximal ideals>, hence pairwise comaximal. The <Chinese remainder theorem> supplies $a\in\mathcal O_F$ satisfying $a\equiv0\pmod{\mathfrak r}$ and $a\equiv1\pmod{\sigma\mathfrak q}$ for every $\sigma\in G$. Its <field norm> is
$$
b=N_{F/K}(a)=\prod_{\sigma\in G}\sigma(a)\in\mathcal O_K.
$$
The factor $a$ puts $b$ in $\mathfrak r$, hence in $\mathfrak r\cap\mathcal O_K=\mathfrak p$. But every $\sigma(a)$ is congruent to $1$ modulo $\mathfrak q$, since $a\equiv1\pmod{\sigma^{-1}\mathfrak q}$. Thus $b\equiv1\pmod{\mathfrak q}$, contradicting $b\in\mathfrak p\subset\mathfrak q$. This proves <transitivity of the Galois action on primes> without any unramified hypothesis.
Let $D=\operatorname{Stab}_G(\mathfrak q)$ be the <decomposition group> and $H=\operatorname{Gal}(F/L)$. Transitivity identifies the primes of $F$ above $\mathfrak p$ with the <coset> space $G/D$. Two such primes contract to the same prime in $L$ exactly when they are in the same $H$-orbit: apply the transitivity just proved to the <Galois extension> $F/L$. Therefore <primes in an intermediate field as double cosets> gives
$$
\boxed{\#\{\mathfrak l\subset\mathcal O_L:\mathfrak l\mid\mathfrak p\}=|H\backslash G/D|=|D\backslash G/H|.}
$$
The second equality follows by inversion of representatives. It allows one to count $D$-orbits on the embeddings of $L$ into $F$, represented by $G/H$. If $H$ is normal, this simplifies to $[G:HD]$; the double-coset formulation also handles nonnormal intermediate fields.
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