This is closely related to the class-field-theoretic type argument I mentioned in the other answer. Here's a fast solution (if possibly not the most elementary).
Note that by class field theory, specifically properties of the Artin symbol, the order $m$ of $[p]$ equals the relative inertia degree of $p$ in the Hilbert class field $K^{(1)}$ of $K$, and similarly for the order of $[\mathcal{P}]$ and the residue degree of $\mathcal{}P$ in $L^{(1)}/L$. Now consider the following diagram:
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} L & \ra{m':=\text{ order of }\mathcal{P}} & L^{(1)} \\ \da{f:=f(\mathcal{P}/p)} & & \da{}\\ K & \ra{m :=\text{ order of }p} & \,K^{(1)} \\ \end{array}
Here the labels on each arrow are the inertia degrees of a prime above $p$ in each extension. Now by multiplicativity of inertia degrees in towers, we have m\mid m'f, so \frac{m}{(m,f)}\mid m', as desired.
(Apologies for the sloppy diagram, I've never jury-rigged an xypic on MSE or MO before. Can anyone make those arrows "headless"? And/or move the$f(\mathcal{P}/p)$ to the left of the arrow?)