= Solution
The <decomposition group> of the chosen prime is its stabilizer under the <Galois group>:
$$
\boxed{D=D_{\mathfrak P\mid\mathfrak p}=\{\sigma\in G:\sigma(\mathfrak P)=\mathfrak P\}.}
$$
The <transitivity of the Galois action on primes> identifies the primes of $M$ above $\mathfrak p$ with $G/D$ through $gD\mapsto g\mathfrak P$.
Put $H=\operatorname{Gal}(M/L)$. Contraction of prime ideals to $\mathcal O_L$ is constant on every $H$-orbit, because every element of $H$ fixes $L$ pointwise. It is surjective onto the primes of $L$ above $\mathfrak p$, since prime ideals extend in the integral extension $\mathcal O_M/\mathcal O_L$. Moreover, two primes of $M$ contracting to the same prime of $L$ are conjugate by $H$: the extension $M/L$ is Galois and the same transitivity theorem applies to it.
Thus the fibers of contraction are exactly the $H$-orbits on $G/D$. These orbits are the <double cosets> $H\backslash G/D$, proving the <primes in an intermediate field as double cosets> correspondence
$$
\boxed{H\backslash G/D\xrightarrow{\sim}\{\mathfrak q\subset\mathcal O_L:\mathfrak q\mid\mathfrak p\},\qquad
HgD\longmapsto g\mathfrak P\cap\mathcal O_L.}
$$
Explicitly, replacing $g$ by $hgd$ does not change the contracted prime. Conversely, if $g\mathfrak P$ and $g'\mathfrak P$ contract to the same prime, some $h\in H$ sends the former to the latter, so $g'^{-1}hg\in D$ and $Hg'D=HgD$. This also verifies the stated order of the two sides of the double coset.
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