Let the perpendicular to $PA$ at $P$ cut $AC,AB,BC$ at $Y,Z,T^*,$ resp and let $X \equiv RZ \cap QY.$ Since $AC,AB$ are perpendicular bisectors of $PQ,PR,$ then $A$ is the center of $\odot(PQR)$ $\Longrightarrow$ $YZ,$ $QY,$ $RZ$ are tangents at $P,Q,R$ $\Longrightarrow$ $XP$ is the polar of $T^*$ WRT $\odot(PQR)$ $\Longrightarrow$ $X(Z,Y,P,T^*)=-1$ $\Longrightarrow$ $A(B,C,P,T^*)=-1$ or $P(B,C,A,T^*)=-1.$ As a result, $\angle APT=90^{\circ}$ $\Longleftrightarrow$ $T \equiv T^*$ $\Longleftrightarrow$ $PA$ bisect $\angle BPC$ $\Longleftrightarrow$ $\angle APB=\angle APC.$

Sorry there is a flaw in the previous resolution. $T^*$ is the pole of $XP$ WRT $\odot(PQR)$ only when $T \equiv T^*,$ though this can be easily fixed keeping the same notations. If $\angle APT=90^{\circ}$ $\Longrightarrow$ $T$ is the pole of $XP$ WRT $\odot(PQR)$ $\Longrightarrow$ $X(Z,Y,P,T)=-1$ $\Longrightarrow$ $P(B,C,A,T)=-1$ $\Longrightarrow$ $AP$ bisects $\angle BPC.$ Conversely if $AP$ bisects $\angle BPC$ $\Longrightarrow$ $P(B,C,A,T^*)=-1$ or $A(B,C,P,T^*)=-1$ $\Longrightarrow$ perpendiculars from $P$ to $AB,AC,AP,AT^*$ form a harmonic pencil as well. If $D \in \odot(PQR)$ is the reflection of $P$ on $AT^*,$ then $P(Q,R,D,T^*)=-1$ $\Longrightarrow$ $PQDR$ is harmonic $\Longrightarrow$ $Q,R,T^*$ are collinear $\Longrightarrow$ $T \equiv T^*$ $\Longrightarrow$ $\angle APT=90^{\circ}.$

After performing the Inversion with center $ A $ we get the following equivalent problem : Given a $ \triangle ABC $ and an arbitrary point $ P $. Let $ Q, $ $ R $ be the reflection of $ P $ in $ CA, $ $ AB $, respectively. Let $ T $ be the second intersection of $ \odot (ABC) $ and $ \odot (AP) $. Prove that $ A, $ $ Q, $ $ R, $ $ T $ are concyclic if and only if $ \angle PBA $ $ = $ $ \angle PCA $. ____________________________________________________________ Proof : Let $ Y, $ $ Z $ be the projection of $ P $ on $ CA, $ $ AB $, respectively. Since $ A $ lies on the perpendicular bisector $ CA, $ $ AB $ of $ PQ, $ $ PR $, so $ A $ is the circumcenter of $ \triangle PQR $ $ \Longrightarrow $ $ \angle AQR $ $ = $ $ \angle ARQ $ $ = $ $ 90^{\circ} $ $ - $ $ \angle BAC $, hence $ T $ lies on $ \odot (AQR) $ iff $ \angle QTP $ $ = $ $ \angle RTP $ $ = $ $ \angle BAC $. On the other hand, since $ T $ is the Miquel point of the complete quadrilateral $ \{ BC, CA, AB, YZ \} $, so $ \triangle TYC $ $ \sim $ $ \triangle TZB $ $ \Longrightarrow $ $ \angle QTP $ $ = $ $ \angle RTP $ $ = $ $ \angle BAC $ iff $ \triangle TYC $ $ \cup $ $ (P,Q) $ $ \sim $ $ \triangle TZB $ $ \cup $ $ (R,P) $ (notice $ PQ $ $ \perp $ $ CY, $ $ RP $ $ \perp $ $ BZ $ and $ Y, $ $ Z $ is the midpoint of $ PQ, $ $ RP $, respectively.) iff $ \angle PBA $ $ = $ $ \angle PBZ $ $ = $ $ \angle RBZ $ $ = $ $ \angle QCY $ $ = $ $ \angle PCY $ $ = $ $ \angle PCA $.

Different solution by inversion: Consider an inversion $\Psi$ with center $P$. After performing $\Psi$ we get the following problem: Problem: Let $PBC$ be a triangle. Consider an arbitrary point $A$. Let $R,Q$ be the circumcenters of $\triangle PAB,\triangle PAC$ respectively. Let $T$ be a point on circumcircle of $\triangle ABC$ such that $PT\perp PA$ then $RPTQ$ is cyclic if and if only $\angle APB=\angle APC$. Proof: Let $O$ be the circumcenter of $\triangle PBC$ then $RO,OQ$ are perpendicular bisectors of $PB,PC$ respectively. since $RQ$ is perpendicular bisector of $PA$ so $PT\parallel RQ$. also $OP=OT$. Now if: 1) $RPTQ$ is cyclic then it is isosceles trapezoid so $PR=QT$ and $\angle OPR=\angle OTQ$ hence $\triangle OTQ=\triangle OPR\Longrightarrow OR=OQ\Longrightarrow \angle APC=\angle APB$. 2) $\angle APC=\angle APB$ then $\angle OQR=\angle ORQ\Longrightarrow OR=OQ$ Also since $T$ is midpoint of arc $BPC$ of $\odot(PBC)$ we get $\angle TOQ=\angle POR=\angle C$ hence $\triangle OTQ=\triangle OPR\Longrightarrow TQ=PR\Longrightarrow RPTQ$ is isosceles trapezoid so it is cyclic. DONE

Dose anyone notice that the claim is not true when $P$ lies on the extension of segment $BC$ ? I wonder if the translation is correct.

a bit similar to Luis Gonzalez's solution perhaps: lemma: for two points $P,Q$ on a circle $\omega$ and $A$ another point on $\omega$ then if $(AA,PP)=S$, $(AA,QQ)=T$, $(PQ,AA)=B$ $\implies (STAB)=-1$ Back to problem: let $A'=(AP,BC)$ we easily see that $A \equiv O_{PQR}$ and if then let $T'$ be a point on $BC$ such as that $PT' \perp AP$ so its tangent to $\odot PQR$ and let $PT'$ meet $AB, AC$ respectively at $Z, Y$ then if $PA$ bisects $\angle BPC$ then easy to see $(BCA'T')=(ZYPT')=-1$ and according to the lemma $(ZYPU)=-1$ in which $U$ is the cross point of $ZY$ and $QR \implies U \equiv T' \equiv T$ and the same for the converse !

Let $K$ be a point on $RQ$ such that $\angle{KPA}=90$ and let $M$ midpoint of $PK$ and $L$ foot of perpendicular of $P$ to $AK$ and $E,F$ be the midpoint of $PR , PQ$ . $ALFPE$ is cyclic ($\omega$) and $PK$ is tangent to $\omega$ .Because $MP=ML$ and $MP $ is tangent to $\omega$ , $ML$ is tangent , too . So $(FE,LP)=-1$ , so $A(BC,KP)=-1$ so if we get $X$ and $D$ intersection of $AK , AP$ with $BC$ , we can proof the else easily