= Solution
Use the <Hilbert class field> theorem stated in part (a), and let $H_L$ be the Hilbert class field of $L$. Put $F=M\cap H_L$ inside a common algebraic closure.
Any intermediate extension of $M/L$ is totally ramified at the given prime. To verify this even when $M/L$ is not Galois, there is a unique prime of $F$ above $\mathfrak p$, its residue field embeds into the residue field at the unique prime of $M$ and therefore equals $k_{\mathfrak p}$, and the <fundamental identity for prime decomposition> gives ramification index $[F:L]$. On the other hand $F/L$ is a subextension of $H_L/L$, so it is unramified there. Its ramification index is consequently both $[F:L]$ and $1$, forcing
$$
\boxed{M\cap H_L=L.}
$$
Because $H_L/L$ is Galois, this intersection identity gives
$$
[MH_L:M]=[H_L:L]=h_L,\qquad
\operatorname{Gal}(MH_L/M)\cong\operatorname{Gal}(H_L/L).
$$
Thus $MH_L/M$ is abelian. We use the standard local fact that unramified extensions remain unramified under finite base extension: after base change each factor is obtained by lifting an extension of the residue field. Since $H_L/L$ is unramified at every finite prime, $MH_L/M$ is also unramified at every finite prime. There are no real places of $M$, since every embedding of $M$ restricts to an embedding of the totally imaginary field $L$. Hence there is no infinite-place obstruction.
The maximality of $H_M$ now implies $MH_L\subseteq H_M$. Taking degrees over $M$ proves the <class number divisibility under total ramification>:
$$
\boxed{h_L=[MH_L:M]\mid[H_M:M]=h_M.}
$$
The argument uses the degree of an unramified class-field extension, and does not assume that extension of ideals induces an injection of <ideal class groups>.
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