Let $B'$ be a point such that $BB'\parallel AC$ and $B'\in \odot (ABC)$ similarly defined $C'$, so from $ACB'B$ is an isoseles trapezoid we get $MB$ $=$ $MB'$ $\Longrightarrow$ $BB'$ is the radical axis of $\odot (ABC)$ and $\omega_B$ similarly $CC'$ is the radical axis of $\odot (ABC)$ and $\omega_C$ $\Longrightarrow$ $E=BB'\cap CC'$ belongs in the radical axis of $\omega_B$ and $\omega_C$. On the other hand, so from $ABCD$ is an isosceles trapezoid and $ABEC$ is a parallelogram we get $ED\perp BC$ $\Longrightarrow$ $ED\perp MN$ $\Longrightarrow$ $ED$ is perpendicular to the union of centers of $\omega_B$ and $\omega_C$, but $E$ belongs in the radical axis of $\omega_B$ and $\omega_C$, hence $ED$ is the radical axis of $\omega_B$ and $\omega_C$ $\Longrightarrow$ $P$, $Q$ and $D$ are collinear.

could you post all the problem? Thanks

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Easy to see that $PQ\perp BC$, since $PQ\perp MN$ by radical axis and $MN\parallel BC$. Denote X the intersection of PQ and BC. Since it's on the radical axis we have $XM^2-MA^2=XN^2-NA^2$, so $XM^2-XN^2=AM^2-AN^2$. Now when Y is a variable point on BC, $YM^2-YN^2$ can have a fixed value k in 2 positions on BC, symmetric by the point Z where $ZM=ZN$. WLOG, if we choose $AB<AC$ then X has an unique value, so in order to prove that $P,Q,D$ are collinear, it's enough to prove that $D', X$ coincide, where $D'$ is the projection of D on BC. So we just have to prove that $D'M^2-D'N^2=AM^2-AN^2$ which is just easy computation after we project $M,N,A,D$ onto $BC$. Correct me if I'm wrong, and if anyone wants me to add the rest of the calculation I will, although it's pretty easy.