Let $X$ and $X'$ be the intersections of $(AQB)$ and $(APC)$ with $BC$. Then $\angle AXB+\angle CX'A=\angle APB+\angle CPA=\angle APQ+\angle QPA+\angle PQB+\angle CPQ=$ $\angle APQ+\angle QPA+\frac{1}{2}\angle QPA+\frac{1}{2}\angle AQP=\frac{3}{2}(\angle APQ+\angle QPA)=\frac{3}{2}(\pi-\frac{\pi}{3})=\pi$ Therefore $(APB)$ and $(AQC)$ concur on the point $X$ on $BC$. Now taking a $\sqrt{bc}$ inversion+reflection about the $A$ bisector, the two circumferences are mapped to the lines $CX^*Q^*$ and $BX^*P^*$, with $X^*\in (ABC)$. Therefore, by Pascal's theorem on degenerate hexagon $ABBX^*CC$, the intersection of the two tangents in $B$ and $C$, called $T^*$, is collinear with $P^*$ and $Q^*$. Therefore $T$ (the inverse of $T^*$) lies on all circumferences $(APQ)$. Furthermore, it is well known that this the Humpty point, which lies on the median and on the circumference of diameter $AH$.

Here's my synthetic solution : WLG, let $\triangle ABC$ be oriented in the anticlockwise sense. Let the circumcenter of $\triangle ABC$ be $O$. Let $L$ be the point s.t. $\triangle LBC$ is equilateral and $L,A$ lie on the same side of $\overleftrightarrow{BC}$. Note that $A,B,C,L$ are concyclic. Now, we deduce the following : By $BP = CQ$, $\triangle LPB, \triangle LQC$ are directly congruent. So, $\triangle LPQ,\triangle LBC$ are directly similar $\implies \triangle LPQ$ is equilateral and is oriented anticlockwise. By $BP=PQ=QC$; $LP=PB$ and $LQ=QC$. So, $\overleftrightarrow{PO},\overleftrightarrow{QO}$ are the perpendicular bisectors of $\overline{LB},\overline{LC}$ $\implies\measuredangle POQ = 60^{\circ}$ Now, we have $\measuredangle PAQ = \measuredangle PLQ = \measuredangle POQ = 60^{\circ} \implies A,P,Q,L,O$ are concyclic. So, we have that $\odot APQ\equiv\odot AOL$. Now, let $M$ be the midpoint of $\overline{BC}$. Let $X$ be the $A\mathrm{-Humpty}$ point of $\triangle ABC$ i.e. the projection of the orthocenter of $ABC$ onto the $A\mathrm{-median}$. $\therefore MX\cdot MA = MB^2 = MO\cdot ML\implies X,A,O,L$ are concyclic $\implies X$ lies on $\odot APQ$ (As $\odot APQ\equiv\odot AOL$)