Maybe there's a simpler sol but here's what I come up with. Let $E,F$ be the feet from $A,C$, respectively. WLOG $BA<BC$ (idk if that's necessary or not). Claim: $FPDQ$ is harmonic quadrilateral. Proof. Firstly, since $PQ\parallel DC$, we have $QC=PD=PH$. Similarly, we have $PC=QD=QH$. Thus, $PCQH$ is a parallelogram, so $X=PQ\cap HC$ is midpoint of $PQ$. Hence, since $PD=CQ$, we have $$\angle PFD=\angle CFQ=\angle XFQ$$which means that $FD$ is $F$-symmedian of triangle $FPQ$ as desired. Now, we invert about $B$ with radius $\sqrt{BD\cdot BH}$. Denote the mapping from any point $K$ by $K'$. Note that $P'=BP\cap AH,Q'=BQ\cap AH$. We need to prove that $$180^{\circ}=\angle APB+\angle AQB=\angle BFP'+\angle BFQ'\iff \angle BFQ'=\angle AFP'\iff\angle P'FH=\angle Q'FH.$$Or that $(A,H;P',Q')=-1$ since $\angle HFA=90^{\circ}$. However, this is true as a consequence of the above claim after being inverted. $$F\to A,D\to H, P\to P',Q\to Q'$$We are now done!!

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Inverting about $(BC)$, $H^*$ is the Orthocenter Miquel point ($MH \cap (AH)$) and $P, Q$ are such that $\frac{PD}{PH^*} = \frac{QD}{QH^*} = \frac{MD}{MH^*}$, thus $\frac{DP}{DQ} = \frac{H^*P}{H^*Q} \implies (AH)$, which is centered on $PQ$, must be the Apollonian circle containing the locus of points whose distances from $P, Q$ are in the ratio $DP, DQ$. Note that by angle chasing, the bisectors of $\angle{PDQ}$ and $\angle{DAH}$ intersect on $(AH)$; let this point be $U$. Then $\frac{PU}{QU} = \frac{DP}{DQ}$, so $AU$ bisects $\angle{DAH}, \angle{PAQ} \implies AP, AQ$ are isogonals wrt $\angle{HAD}$. Thus, if $W, V = AH \cap (BC)$ and WLOG $W, P$ lie inside $\triangle{ABC}$, if $P', Q' = AP, AQ \cap (BC)$ then $\widehat{VP'} = \widehat{WQ'} = \widehat{CQ} = \widehat{PD} \implies P'Q' \perp BC$, which suffices by angle chasing in $(BC)$. $\square$

Let us consider the isogonal conjugate $\mathcal{H}$ of $PQ$-bisector in triangle $BPQ$. From one hand it will be the set of points $X$ such that $\angle XPB+\angle XQB=0$ (all angles are oriented). From the other hand, $\mathcal{H}$ is a rectangular hyperbola passing through $B$, $P$, $Q$, $C$ and orthocenter $H$ of the triangle $BPQ$. Thus it passes through the point $A$ — the orthocenter of $BHC$ and so $\angle APB+\angle AQB=0$.

More solutions in Russian here: https://youtu.be/HAQuW-Kl9zY

This was great problem we prove this through the following claim : Claim(i):$H$ is orthocentre of $\Delta BPQ$ Proof:$BH\perp PQ$ (given as $PQ$ is perp bisector of $HD$) and as $PHQD$ is a kite so $\angle PQH = \angle PQD = \angle PBD$ and as $BH\perp PQ$ so we have $QH \perp BP$ and thus claim is proved . Claim(ii):$\square HPCQ$ is a ||gm Proof: As , $BD \perp PQ $ and $BD\perp AC$ hence we have $PQ||DC$ and hence $PQCD$ is a isosceles trapezium.and hence $PD=PH=CQ$ and also note that by claim (i) $PH \perp BQ$ and also $\angle BQC=90^{\circ}$ hence the claim is proved. Claim(iii): $\angle BPH = \angle BQH$ Proof: as $H$ is orthocentre of $\Delta PBQ$ so the claim is trivially true.And we have marked all these equal angles by green colour . Claim(iv) : $\angle AHP = \angle HQP$ Let the projection of $P$ on $BC$ be $X$ so $AH=PX$ and we have $\angle AHP = 90^{\circ}-\angle PBX - \angle BPH $ and as $PD = QC $ so we have $\angle PBX = \angle HBQ $ so we have $\angle AHP = 90^{\circ}-\angle HBQ - \angle BQH = \angle PQH $ and we have marked all this equal angles by green colour. Now note that $\angle HAP = \angle AQP =\angle QAC$ suffices . We translate $\Delta APH$ so that $PH$ goes to $CQ$.Now due to translation $APCA"$ is a gm and so $\angle A"AC = \angle ACP =\angle AHP = \angle A''QC $ .So we have $A''AQC$ to be a cyclic quadrilateral and so $\angle HAP = \angle QA''C = \angle QAC$ Now the final part is very easy Note that $\angle APB + \angle HAP + \angle AHP + \angle BPH = 180 ^{\circ}$ and $\angle AQB = \angle HAP + \angle AHP + \angle BPH$ and so we are done !! $\blacksquare$

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Here is another beautiful solution to this nice problem! Let $P'$ and $A'$ be the reflections of $P$ and $H$ about $M$ (midpoint of $BC$) respectively. $\widehat{APB}=270-\widehat{APC}$ and $\widehat{AQB}=90-\widehat{AQC}$, thus $\widehat{APB}+\widehat{AQB}=180 \iff \widehat{APC}+\widehat{AQC}=180$. Since $\widehat{APC}=\widehat{AP'C}$, we need to prove that $AP'CQ$ is cyclic, or $\widehat{QCA}=\widehat{QP'A}$. Note that $P'A'=PH=PD=CQ$ and $P'A'\parallel PH\parallel CQ$, so $A'P'QC$ is a parallelogram thus $P'Q\parallel A'C$ and since $APCP'$ is also a parallelogram, we have $AP'\parallel PC$, so $\widehat{AP'Q}=180-\widehat{PCA'}$. Since $ABCA'$ is cyclic and $A'C\perp BC$ we get that: $\widehat{PCA'}=90-\widehat{PCB}=90-\widehat{PDB}=\widehat{PDA}=180-\widehat{QCA}$, which means $AP'CQ$ is cyclic, so we are done!