Solution:

Let $J$ be the center of the excircle $\omega$. Then $JE \perp AE$ and $JF \perp AF$ hold, so the Thales circle $\Omega$ of $\overline{AJ}$ passes through $E$ and $F$, and hence also through $P$ and $Q$. Let the ray $AD$ meet $\Omega$ and $\omega$ again at $N$ and $T$, respectively.

Then $JN$ is the perpendicular from $J$ to $DT$, so $N$ is the midpoint of the segment $\overline{DT}$. Hence, by the power of a point theorem,
$$|DM| \cdot |DT| = \frac{1}{2} \cdot |DA| \cdot |DT| = |DA| \cdot |DN| = |DP| \cdot |DQ|,$$
so by the converse of the power of a point theorem, the point $T$ lies on the circumcircle of triangle $MPQ$. We now consider the image of $N$ under inversion in the circle $\omega$, and denote it by $Z$. Since the Thales circles over $\overline{JD}$ and $\overline{JT}$ both pass through the point $N$, $Z$ lies on the images of these circles, namely the tangents to $\omega$ through $D$ and $T$. On the former line also lie $B$, $C$, $P$ and $Q$. Furthermore, $N$ lies on the circle $\Omega$, which under the inversion considered is mapped to the line $EF$, so $Z$ also lies on the line $EF$. Hence, by the power of a point theorem (secant form),
$$|ZT|^2 = |ZE| \cdot |ZF| = |ZP| \cdot |ZQ|,$$
so by the converse of the power of a point theorem (secant form), $ZT$ is also tangent to the circumcircle of triangle $MPQ$. Thus we have proven that this circle and $\omega$ are tangent to each other at the point $T$.