Not an answer, but too long to fit into a comment. Clearly the assumption, $H=1$, implies that
the field of fractions $F$ of the intermediate ring $C$ is all of $L$, for otherwise
$Gal(L/F)$ would be a non-trivial subgroup of $H$.
Assume that $\mathfrak{P}=\mathfrak{P}_B\cap C$ for some prime ideal $\mathfrak{P}_B$ (necessarily lying over $\mathfrak{p}$). Let $G_T=G_T(\mathfrak{P}_B\vert\mathfrak{p})$ be the corresponding inertia group:
$$
G_T=\{\sigma\in G\mid \sigma(\mathfrak{P}_B)=\mathfrak{P}_B,\ \forall\,x\in B: \sigma(x)\equiv x\pmod{\mathfrak{P}_B}\}.
$$
Assume that $\sigma\in G_T$. Let $x\in C$ be arbitrary. Then $\sigma(x)-x\in C$ and
$\sigma(x)-x\in\mathfrak{P}_B$, so $\sigma(x)-x\in C\cap\mathfrak{P}_B=\mathfrak{P}$.
Therefore $\sigma\in H$.
As we work under the assumption that $H$ is trivial, we can conclude that $G_T=1.$ Therefore
$$
e(\mathfrak{P}_B\vert \mathfrak{p})=|G_T|=1.
$$
So $\mathfrak{P}_B\vert \mathfrak{p}$ is unramified, and the claim follows (by Galois theory of number fields the ramification indices over all the primes extending $\mathfrak{p}$ are equal).
So the question is reduced to asking whether all the primes of $C$ lying over $\mathfrak{p}$ are gotten by restricting a prime of $B$. The mapping from the primes of $B$
to those of $C$ is not necessarily injective, but the question is about its surjectivity.
Edit: Makoto Kato points out that as $B$ is the integral closure of $C$, the surjectivity follows from general theory.
As an example of the case, where this mapping is not injective I proffer
$K=\mathbb{Q}$, $L=K[\sqrt3]$, $\mathfrak{p}=13$, $C=\mathbb{Z}[13\sqrt3]$. The prime $13$
splits in $L$ as
$$
13=(4+\sqrt3)(4-\sqrt3),
$$
but both prime ideals, $\langle4+\sqrt3\rangle$ and $\langle4-\sqrt3\rangle$ of $B$ intersect $C$ in
$$
\mathfrak{P}=\{a+b\cdot13\sqrt3\mid a,b\in\mathbb{Z},\ a\equiv0\pmod{13}\}.
$$
Anyway, my (possibly confused) thinking here is that $G_T$ would always inject into $H$. For an automorphism to be in $G_T$ we are to some extent asking more (in comparison to $H$), so if we can ascertain a trivial inertia group at the level of $C$, we should expect a trivial inertia group at the level of $B$ as well.
You also need that $\mathfrak{p}$ is unramified. For example, if $K = \mathbf{Q}(i)$ and $\mathfrak{p} = (1+i) \subset \mathbf{Z}[i] = \mathcal{O}_K$, then $(2) = \mathfrak{p}^2$ and $\mathcal{O}_K/\mathfrak{p}^2$ has characteristic $2$, not $4$.
But this is the only other condition that is required. Use the following steps:
- There is an element $y \in \mathcal{O}_K$ such that $y \in \mathfrak{p}$ but not in $\mathfrak{p}^2$.
- There is an isomorphism of groups $\mathbf{F}_p = \mathcal{O}/\mathfrak{p} \rightarrow \mathfrak{p}^{n-1}/\mathfrak{p}^n$ by sending $x \mapsto x y^{n-1}$.
- The order of $\mathcal{O}/\mathfrak{p}^{n}$ is $p^n$.
- (The one step that requires unramified): $p^{n-1}$ is non-zero in $\mathcal{O}_K/\mathfrak{p}^{n}$, because there is a factorization:
$$p^{n-1} = \mathfrak{p}^{n-1} \times \cdots $$
and hence $p^{n-1} \notin \mathfrak{p}^n$.
- Any ring of order $p^n$ were $p^{n-1} \ne 0$ is isomorphic to $\mathbf{Z}/p^n \mathbf{Z}$.
Best Answer
Given a non-zero prime ideal $\mathfrak{p}$ of $O_K$
The first step is to prove that $\mathfrak{p}$ contains an integer, namely the constant coefficient of the minimal polynomial of any of its elements.
Since $\mathfrak{p}$ contains an integer then $O_K/\mathfrak{p}$ has finite characteristic, since it is an integral domain it has characteristic $p$ prime.
Since $K$ is a $n$ dimensional $\Bbb{Q}$- vector space and $O_K$ is a subgroup then $O_K/pO_K$ has at most $p^n$ elements, ie. $O_K/\mathfrak{p}$ is a finite group thus a finite integral domain thus a finite field. Being also a $\Bbb{F}_p$ vector space it has $p^k$ elements.
Up to isomorphism there is only one finite field with $p^k$ elements (the splitting field of $x^{p^k}-x\in \Bbb{F}_p[x]$) thus $O_K/\mathfrak{p}\cong \Bbb{F}_{p^k}$.