Let $O$ be the centre of the circle and $R$ the symmetric of $P$ with respect to $AC$ . $$\angle AOP = 180 - \angle ABC$$so $\angle AQP = \angle AQR = \frac{180 - \angle ABC}{2} $ Now note that $$\angle APB = \angle CPX = \angle PQA = \angle XQC $$so $PQCX$ ia cyclic . The conclusion follows from $$\angle QPC = \angle QXC = \angle QRC $$so $ CR = CP = CX $

Note that $\angle PQC=C-\left(90^{\circ}-\tfrac{1}{2}B\right)+A=90^{\circ}-\tfrac{1}{2}B$ so $\angle PQX=\angle CBA$ and $\angle XPQ=\angle CAB$ proving $\triangle XPQ \sim \triangle CAB$. So $\angle PXQ+\angle PCQ=180^{\circ}$ proving $PCQX$ cyclic. Since $QC$ bisects angle $PQX$ externally, we're done.

1.Chase angles with tangent secent. 2. show that quadrilateral $CXPQ$ is concyclic. 3.Then prove equality of two angle.

Diagram+Soluton

Let R be the second intersection of QX with the circle. Angle AQP=Angle AQR=Angle CQX ....(1) Angle AQP=Angle BPA=Angle CPX ....(2) From (1) &(2) Angle CQX=Angle CPX Thus CQPX is cyclic. Hence, Angle PQA=Angle PXC .... (3) From (2)&(3) Angle PXC=Angle CPX Thus PC=CX

Let P' be reflection of P across AC. ∠XQC = ∠P'QA = ∠AQP = ∠APB = ∠XPC ---> QCXP is cyclic. ∠CXQ = ∠CPQ = ∠CP'Q ---> CX = CP' = CP. we're Done.