= Solution
A prime is completely split in an extension of degree $n$ exactly when there are $n$ primes above it. Indeed, the <fundamental identity for prime decomposition> is $\sum_{\mathfrak q\mid\mathfrak p}e_{\mathfrak q}f_{\mathfrak q}=n$, and every summand is a positive integer. Thus $n$ primes means that every ramification index and residue degree equals one, and conversely.
By part (a), the number of primes of $L$ above $\mathfrak p$ is the number of right $D$-orbits on $H\backslash G$. This set has $[G:H]=[L:K]$ elements. It has that many orbits exactly when every orbit is a singleton, equivalently when
$$
HgD=Hg\quad\text{for all }g\in G.
$$
This condition says $gDg^{-1}\subseteq H$ for every $g$, or
$$
D\subseteq\bigcap_{g\in G}g^{-1}Hg=\operatorname{Core}_G(H),
$$
the <subgroup core>.
Because $M$ is the <Galois closure> of $L/K$, it is generated by all the fields $g(L)$. The fixed field of $gHg^{-1}$ is $g(L)$, so the fixed field of the core is their compositum, namely $M$. The <Galois correspondence> therefore gives $\operatorname{Core}_G(H)=1$. Hence
$$
\mathfrak p\text{ completely split in }L\quad\Longleftrightarrow\quad D=1.
$$
In the Galois extension $M/K$, the number of primes above $\mathfrak p$ is $[G:D]$, so $D=1$ is equivalent to complete splitting in $M$. This proves that <complete splitting is preserved by Galois closure>:
$$
\boxed{\mathfrak p\text{ completely split in }L\quad\Longleftrightarrow\quad
\mathfrak p\text{ completely split in }M.}
$$
No prior unramifiedness assumption is needed: complete splitting already forces all ramification indices to be one, and the decomposition-group argument includes the ramified case.
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