jmc

We start with a diagram, including median $\overline{QN}$, which is also an altitude. Let the medians intersect at $G$, the centroid of the triangle.



We have $NP=PR/2=16$, so right triangle $PQN$ gives us $$\begin{array}{rl}QN& =\sqrt{P{Q}^2-P{N}^2}=\sqrt{34^2-16^2}\\ & =\sqrt{(34-16)(34+16)}=30.\end{array}$$ (We might also have recognized that $PN/PQ=8/17$, so $QN/PQ=15/17$.)

Since $G$ is the centroid of $△PQR$, we have $GN=\frac{1}{3}(QN)=10$, and right triangle $GNP$ gives us $$GP=\sqrt{G{N}^2+N{P}^2}=\sqrt{100+256}=2\sqrt{25+64}=2\sqrt{89}.$$ Finally, since $G$ is the centroid of $△PQR$, we have $PM=\frac{3}{2}(GP)=\boxed{3\sqrt{89}}$.