Denote by $S$ the midpoint of $PQ$ and $T$ the midpoint of $PR$. Because $AP=AQ $ and $BP=BR$ we find that $ [AS$ is the bisector of $\angle{PAQ}$ and $[BT$ is the bisector of $\angle {PBR}$. $m(\angle STP)=m(\angle QRP)=m(\angle ARQ)-m(\angle BRP)= 90^{\circ}-\frac{m(C)}{2}-\frac{m(B)}{2}=\frac{m(A)}{2} \Rightarrow m(\angle STB)=90^{\circ}+\frac{m(A)}{2}$ $m(\angle SAB)= 90^{\circ}-\frac{m(A)}{2}$. We obtain that $m(\angle SAB)+m(\angle STB)=180^{\circ}$, so the quadrilateral $SABT $ is cyclic. q.e.d

Following the same idea, i.e. to prove that $ASTB$ is cyclic looks much easier this way: $AS$ and $BT$ concur at $I_a$, the $A$ -excenter, so $I_aQ\perp AC$ and $I_aR\perp BC$; also $I_aQ=I_aR$; consequently, $I_aQ^2=I_aS\cdot I_aA=I_aT\cdot I_aB=I_aR^2$ and, indeed, $ASTB$ is cyclic. Best regards, sunken rock

We consider the $A-$excircle. let $I_A$ be the $A-$excenter, $P$, $Q$, $R$ be the touchpoints of the $A-$excircle with $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, respectively. Let $M$, $N$ be the midpoints of $\overline{PR}$, $\overline{PQ}$, respectively. Then obviously $B$, $M$, $I_A$ are collinear and $\overline{I_AP} \perp \overline{BC}$, $\overline{PM} \perp \overline{BI_A} \implies I_AM \cdot I_AB=IP^2$. Similarly $I_AN \cdot I_AC=IP^2$. Thus $MNBC$ is cyclic and $ABCM$ is cyclic $\iff$ $ABCN$ is cyclic, as desired.