Using the property of complete quadrilateral we have BQPC concyclic.Because ADP=C-B so only prove QBP=C-B FPB=x then we can easily have PBC=x+B-C HBC=x which is exactly what we are gonna prove.

Let $EF$ meet $BC$ at $T$.Since,$(C,B;D,T)=-1$ and $DP \perp EF,DP$ must be the internal bisector of $\angle BPC$.As $Q$ lies on the perpendicular bisector of $BC,Q$ must be the midpoint of arc $BPC$.Hence,$B,P,Q,C$ are concyclic.By angle chase,$\angle ADP=\angle B-\angle C$.Let $\angle BPF=\alpha$.Then,$\angle BCQ=\angle CBQ=\alpha,\angle BPD=90^{\circ}-\alpha,\angle BDP=90^{\circ}+\angle C-\angle B$.All these imply $\angle PBQ=\angle B-\angle C$ and we are done.

My solution: WLOG assume that $AB \leq AC$. Let $M$ be the midpoint of $BC$, and let $EF \cap BC=X$. Note that $PDMQ$ is cyclic. Using the converse of the Lemma here (which can be proved in a similar way as the proof of the Lemma itself in this link), we get $XB \cdot XC=XD \cdot XM=XP \cdot XQ$, giving that $BPQC$ is cyclic. As $QB=QC$, this gives that $PQ$ is the external angle bisector of $\angle BPC$. Now, Let $\ell$ be the line through $Q$ perpendicular to $EF$. Then as $\ell \parallel DP$ and $QM \parallel DA$, we get that $$\angle ADP=\angle (\ell,QM)=90^{\circ}-\angle PQM=90^{\circ}-\angle BQM-\angle PQB=\angle CBQ-\angle PQB$$And, $\angle CBQ=\angle CPQ=\angle FPB$. Thus $$\angle PBQ=\angle FPB-\angle PQB=\angle ADP \text{ } \blacksquare$$

From Harmonic Bundle argument, obviously $PD$ is angle bisector of $\angle BPC$. Let $\odot (BPC)$ $\cap$ $EF$ $=$ $Q'$, then, $Q'$ is the midpoint of arc $BPC$ $\implies $ $Q' \equiv Q$. Let $PD$ $\cap$ $\odot (BPC)$ $=$ $D'$ $\implies$ $\angle ADP$ $=$ $\angle PD'Q$ $=$ $\angle PBQ$

Maybe same as others FBH 2018 TST Day 1 P1 wrote: In acute triangle $ABC$ $(AB < AC)$ let $D$, $E$ and $F$ be foots of perpedicular from $A$, $B$ and $C$ to $BC$, $CA$ and $AB$, respectively. Let $P$ and $Q$ be points on line $EF$ such that $DP \perp EF$ and $BQ=CQ$. Prove that $\angle ADP = \angle PBQ$ Let $M$ be the midpoint of $BC$. So $QM\perp BC$. Hence, $D,P,Q,M$ are concyclic. Let $EF\cap BC=X_A$. Then as $(X_AD;BC)=-1$. Hence, by Maclaurin's Theorem we get that $$X_AB\cdot X_AC=X_AD\cdot X_AM=X_AP\cdot X_AQ\implies B,P,Q,C\text{ are concyclic.} $$Let $QM\cap PD=T$. $PD$ is the internal bisector of $\angle BPC$ as $(X_AD;BC)=-1$ and $\angle DPX_A=90^\circ$. So, $P,B,T,C$ are also conyclic. Hence, $P,Q,C,T,B$ are all concyclic. So, $$\angle ADP=\angle QX_AC=\angle PTQ=\angle PCQ=\angle PBQ.\blacksquare$$

Firstly, let's denote $M$ as midpoint of $BC$ and let $\ell$ be perpendicular bisector of $BC$. Then we know that $Q = \ell \cap EF$. So from there $\angle QMD = \angle QPD = 90^\circ$ hence $D, P, M, Q$ are concyclic. Let $G = EF \cap BC$. Now lets denote circumcircle of $DPMQ$ as $\omega_1$. Now we have $$Pow_{\omega_1}(G) = GP \cdot GQ = GD \cdot GM$$We want to prove that $B, P, C, Q$ are concyclic. To do that we are left to prove $$GD \cdot GM = GB \cdot GC$$Which follows from Manclaurin's theorem because $(GD;BC) = - 1$. From this we conclude that $BPCQ$ is cyclic. Easily we obtain $\angle APD = \angle B - \angle C$. Let $BPF = x$. Then $angle BCQ = \angle CBQ = x$, and so $$\angle BPD = 90^{\circ} - x, \angle BDP = 90^circ - \angle B + \angle C$$This implies $\angle PBQ = \angle B - \angle C$ thus we are done.