Solution:

Let us denote by $U$ the center of the circumscribed circle of $\triangle BME$. Let us apply inversion with center $A$ and square of radius $AB \cdot AC$. Points $B$ and $C$ map to points $B'$ and $C'$ symmetric to points $C$ and $B$ with respect to $AP$, points $P$ and $M$ map to each other, and $E$ maps to a point $E'$ symmetric to $Q$ with respect to $AP$. Therefore, the line $AU$ coincides with the line joining $A$ with the center of the circle $B'PE'$ (of course, the centers do not map to each other!). We see that this line is symmetric to

the line $AZ$ with respect to the bisector of angle $A$, where $Z$ is the center of the circle circumscribed about $\triangle CPQ$.

Analogously we obtain that the line $BZ$ is symmetric to the line joining $B$ with the center $V$ of the circle $AND$ with respect to the bisector of angle $B$. By Ceva's theorem in trigonometric form (or by the statement about isogonally conjugate points), the lines symmetric to the lines $AU, BV, CX$ with respect to the bisectors of angles $A, B, C$ respectively also meet at one point, which means that the line $CZ$ is symmetric to $CX$ with respect to the bisector of angle $C$. But $Z$ is the center of the circle $CPQ$, from which it follows that the line $CX$ contains the altitude of triangle $CPQ$, and that is what we wanted to prove.