Clearly, $ (P,Q) $ are isogonal conjugatets, so, if we draw $ QN \perp BC $ and $ QM\perp AB $, then ,clearly $ M,F,E,D,N $ lies on the same circle, with mid point of $ PQ $ is it's center, but also note that as $ \angle FED=90^0 $ ,it implies the center lies on the mid point of $ DF $ , hence we conclude that $ PQ $ and $ DF $ bisects each other, so, $ FQDP $ is a parallelogram, hence $ FQ\parallel PD\implies FQ\perp BC $ and also $ DQ\parallel PF\implies DQ\perp AB $ , hence we are done!

Why $M, N, D, E, F$ are concyclic ? And why the circle's center is the midpoint of $PQ$ ?

For details , see EPISODES IN NINETENTH AND twentieth century euclidean geometry by ROSS HONSBERGER in the chapter 'on symmedian points '

The theorem about pedal circles in this blog post trivializes the problem, but that's already been noted above. Here's a solution that doesn't require knowledge of isogonal conjugates, so it's hopefully acceptable now for me to put this on handouts. We will show that $Q'$, the orthocenter of $\triangle BDF$, satisfies the conditions of $Q$. To begin, observe that $DQ'FP$ is a parallelogram, as $P$ is the antipode of $B$ with respect to the circumcircle of $\triangle BDF$. Furthermore, $$\angle DPC = \angle DEC = 90^\circ - \angle FEA = 90^\circ - \angle FPA,$$which, in conjunction with $\angle PDC = \angle AFP = 90^\circ$, yields $\triangle CPD\sim\triangle PAF$. Combining these two results, we have $$\frac{CD}{DQ'} = \frac{CD}{PF} = \frac{CP}{PA}.$$ [asy][asy] size(7.5cm); defaultpen(fontsize(9pt)); pair A=(57,142.5), B=(0,0), C=(207.3,0), D=(100,0), E=(112.8,89.6), F=(40,100), P=(100,76), Q=(40,24); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(P); dot(Q); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, NE); label("$F$", F, NW); label("$Q'$", Q, SW); label("$P$", P, SW); draw(A--B--C--cycle, linewidth(0.5)); draw(P--D, linewidth(0.5)); draw(P--E, linewidth(0.5)); draw(P--F, linewidth(0.5)); draw(D--E--F--cycle, linewidth(0.5)); draw(F--Q--D, linewidth(0.5)); draw(A--Q--F--cycle, linewidth(1.1)); draw(C--Q--D--cycle, linewidth(1.1)); draw(A--P--C--cycle, linewidth(1.1)); [/asy][/asy] From here, directly angle chase to obtain \begin{align*} \angle CDQ' &= 180^\circ - \angle BDQ' \\ &= 180^\circ - (90^\circ - \angle B)\\ &= 90^\circ + \angle B\\ &= \angle PED + \angle PEF + \angle B\\ &= \angle PCB + \angle PAB + \angle B\\ &= \angle CPA, \end{align*}from which it follows that $\triangle CDQ'\sim\triangle CPA$, so $\angle ACP=\angle BCQ$. Symmetry implies that $\angle CAP = \angle BAQ$ as well, so $Q=Q'$, and we're done.