Very nice problem Let $B', C'$ be the circumcenters of $(APB), (APC)$ and redefine $Q$ as the isogonal conjugate of $P$ in $\triangle ABC$. Let $X,Y$ be the intersections of the perpendicular bisector of $PQ$ with $\Omega$. It suffices to show $B', C', X, Y$ are concyclic. Let $D$ be the circumcenter of $\triangle APQ$ and $E = AD\cap \Omega$. Clearly $D\in XY, B'C', AE$ so it suffices to show $DX\cdot DY = DB'\cdot DC'$, or equivalently that $AEB'C'$ is cyclic. Note that $AE$ is isogonal to the $A$-altitude in $\triangle APQ$, and $\angle PAQ, \angle BAC$ share the same angle bisector. Also note that if $P,Q$ are isogonal conjugates in $\triangle ABC$ before $\sqrt{bc}$-inversion, then they are afterwards as well. Therefore, after applying $\sqrt{bc}$-inversion, it suffices to show the following: If $P,Q$ are isogonal conjugates in $\triangle ABC$ and $A_1, A_2$ are the reflections of $A$ in $CP,BP$, then $R = A_1A_2\cap BC$ satisfies $AR\perp PQ$. First we will symmetrize the diagram by defining $A_3, A_4$ as the reflections of $A$ in $BQ, CQ$. Then it suffices to show that $R = A_1A_2 \cap A_3A_4$ lies on $BC$ with $AR\perp PQ$. However, it's clear from angle-chasing that $\angle A_2BC = \angle B - 2\angle ABP = \angle PBQ = \angle A_3BC$, and $BA_2=BA_3$, so $A_2, A_3$ are reflections across $BC$. Similarly, $A_1, A_4$ are reflections across $BC$ as well, so $R$ trivially lies on $BC$. Furthermore, $A_1A_2A_3A_4$ is an isosceles trapezoid, so it must be cyclic, hence by radical axis on $(A_1A_2A_3A_4), (AA_1A_2), (AA_3A_4)$ we know $AR \perp PQ$ as desired.

Solved with CyclicISLscelesTrapezoid. Let $O_A, O_B, O_C$ be the centers of $(PBC), (PAC), (PAB)$ respectively, $Q$ be the isogonal conjugate of $P$ wrt $\triangle{ABC}$, and $Q_A, Q_B, Q_C$ be the centers of $(QBC), (QAC), (QAB)$. Claim 1: $(O_AO_BO_C), (Q_AQ_BQ_C), (ABC)$ are coaxial. Proof: If $Q_A$ is the center of $(QBC)$, then $O_A$ and $Q_A$ lie on the perpendicular bisector of $BC$, and $\angle{BO_AC} + \angle{BQ_AC} = 2(180^{\circ} - \angle{A})$, so $O_A$ and $Q_A$ are inverses wrt $(ABC)$. Thus, inverting about $(ABC)$ proves the claim. Let their radical axis intersect $(ABC)$ at $X$ and $Y$. Claim 2: $\triangle{PXY} \sim \triangle{QXY}$. (This would solve the problem.) Proof: Invert about $P$. Then $\angle{ABQ} = \angle{CBP} = \angle{PO_AO_B} = \angle{PO_B^*O_A^*} \implies \triangle{O_A^*O_B^*O_C^*} \sim \triangle{ABC}$, and $P$ has the same orientation wrt $\triangle{O_A^*O_B^*O_C^*}$ as $Q$ does wrt $\triangle{ABC}$. Furthermore, $A^*, B^*, C^*$ wrt $P$ and $\triangle{O_A^*O_B^*O_C^*}$ are the same as $Q_A, Q_B, Q_C$ wrt $Q$ and $\triangle{ABC}$, so $\triangle{PY'X'} \sim \triangle{QXY}$. $\triangle{PY'X'} \sim \triangle{PXY}$, so we are done. $\square$

Oh my gosh. This problem is so cool. Let $Q$ be the isogonal conjugate of $P$. Let $O$ be the center of $(ABC)$. Let $P_A$ be the center of $(PBC)$ and similarly define $P_B$, $P_C$, $Q_A$, $Q_B$, $Q_C$. Define $O_A$ as the center of $(APQ)$ and similarly define $O_B$ and $O_C$. Claim 1: $OQ_B \cdot OP_B=OA^2$ Obviously $OP_BQ_B$ are collinear along the bisector of $AC$. It suffices to check $\angle OAQ_B = \angle OP_BA$ and then we would have $OAQ_B ~ OP_BA$, from which the conclusion follows. To do this, we angle chase, letting $f(RS)$ be the direction of line $RS$ in degrees, taken modulo $180$. Recall that the formula for the direction of $OA$ where $O$ is the circumcenter of $(ABC)$ is given by $f(OA)=f(AB)+f(AC)-f(BC)+90$ (a simple proof of this fact is the orthocenter and circumcenter being isogonal). We now compute $$f(AP_B)+f(AQ_B)=(f(AC)+f(AP)-f(PC)+90)+(f(AC)+f(AQ)-f(CQ)+90)=2f(AC)+(f(AP)+f(AQ))-(f(CP)+f(CQ))$$$$=2f(AC)+(f(AB)+f(AC))-(f(AC)+f(BC))=2f(AC)+f(AB)-f(BC)=90+f(OA)+f(AC)=f(P_BQ_BO)+f(OA)$$This implies the desired angle equality. We now finish the problem. From the claim, we symmetrically get $OQ_A \cdot OP_A=OA^2$ and $OQ_C \cdot OP_C=OA^2$. This implies that $P_AP_BQ_AQ_B$, $P_BP_CQ_BQ_C$, and $P_AP_CQ_AQ_C$ are all cyclic by power of a point from $O$. We also know that inversion around $O$ with radius $OA$ swaps $(P_AP_BP_C)$ with $(Q_AQ_BQ_C)$, implying these circles are coaxial. Now observe that $O_A=P_BP_C \cap Q_BQ_C$. From $P_BP_CQ_BQ_C$ cyclic we get $O_AP_B \cdot O_AP_C = O_AQ_B \cdot O_AQ_C$, so $O_A$ is on the common radical axis. Similarly $O_B$ and $O_C$ are on the radical axis. Since $O_A$, $O_B$, and $O_C$ are all on the perpendicular bisector of $PQ$, $P$ must be the reflection of $Q$ over the common radical axis.

Very beautiful problem! Redefine $Q$ to be the isogonal conjugate of $P$ wrt triangle $ABC$. Let $O$ be the center of $(ABC)$. Let $P_A,Q_A$ denote the center of $(PBC),(QBC)$ respectively. Similarly define $P_B,Q_B,P_C,Q_C$. Let $\omega = (ABC), \omega_P = (P_AP_BP_C),\omega_Q= (Q_AQ_BQ_C)$. Claim 1 : $(P_A,Q_A),(P_B,Q_B),(P_C,Q_C)$ are inverses wrt $\omega$ and hence $\omega,\omega_P,\omega_Q$ are coaxial. Proof : By angle chase we can show that $\angle BPC + \angle BQC = 180+A$. Using this its not hard to show that $\angle OBP_A = \angle OBC + \angle CBP_A = \angle OQ_AB$. This proves that $P_A,Q_A$ are inverses wrt $\omega$. Now invert about $\omega$ and the coaxiality follows. Let $\ell$ denote the common radical axis of the 3 circles. Claim 2 : Circumcenters of $(APQ),(BPQ),(CPQ)$ lie on $\ell$ and hence $\ell$ is perpendicular bisector of $PQ$. Proof : From claim 1 we can infer that $P_BQ_BP_CQ_C$ is cyclic quadrilateral. By radax on $\omega_P,\omega_Q,(P_BQ_BP_CQ_C)$ we infer that $\ell, P_BP_C,Q_BQ_C$ concur. But observe that $P_BP_C \cap Q_BQ_C$ is magically the circumcenter of $APQ$! This proves the claim! From claim 1 and claim 2, we are done

Nice problem! The idea to exploit symmetry with inversion works well here! Redefine $Q$ to be the isogonal conjugate of $P$ WRT $ABC$ Let $D,E,F$ denote the centers of $(BPC),(APC),(APB)$ and let $G,H,I$ denote the centers of $(BQC),(AQC),(AQB)$ Claim 1: $(DEF),(GHI),(ABC)$ are coaxial

Claim 2: The radical center of $(DEF),(EFHI),(GHI)$ is circumcenter of $(APQ)$

By Claim 2, we get $XY$ is perpendicular bisector of $PQ$ hence we're done!