Without loss of generality assume that $P$ lies between $A$ and $Q$. Let the line passing through $M$ parallel to $AC$ intersect the sides $AB$ and $BC$ at the points $K$ and $L$ respectively. Since $M$ is the midpoint of the arc $PQ$, the line $KL$ is parallel to $AC$ whence $\omega$ is the tangency point of $B$-excircle of the triangle $BKL$. Thus the point $X = BM \cap AC$ is the tangency point of $B$-excircle of the triangle $BAC$. Therefore
$$2AX = AC + BC - AB = AC + \sqrt{CP \cdot CQ} - \sqrt{AP \cdot AQ},$$
we used the fact that the difference $BC - AB$ equals to the difference between the lengths of the tangents from $C$ and $A$ to $\omega$.
Note that $2AX$ doesn't depend on $\omega$, this circle must only pass through $P$ and $Q$. Therefore if $Y = DM \cap AC$ then $2AX = 2AY$, i.e. $X = Y$ and $BM$, $DN$ and $AC$ are concurrent.