Let E be a point such that $PE=AP,QE=BQ$($C,E$ are on the same side regarding $PQ$). Since $\frac {PE}{PQ}=\frac{AP}{PQ}=\frac{BC}{AB}$, and $\frac {QE}{PQ}=\frac{AC}{AB}$, By SSS similarity we deduce $\bigtriangleup EPQ \sim \bigtriangleup CBA$ $\Rightarrow$ $\angle PEQ=\angle QCP$ $\Rightarrow$ $C,P,Q,E$ are concyclic $\angle APE=\angle EQB$. Since $AP=PE, QE=QB$, So $\bigtriangleup EPA \sim \bigtriangleup EQB$ $\Rightarrow$ $\bigtriangleup EPQ \sim \bigtriangleup EAB$ $\Rightarrow$ $\angle AEB=\angle PEQ=\angle ACB$ $\Rightarrow$ $C,A,B,E$ are concyclic. Let $K$ be the center of cyclic quad. $CEQP$. Since K,O are two homologous points wrt the spiral similarity transformation that sends $%Error. "bigtriangle" is a bad command. EPQ$ to $\bigtriangleup EAB$. So we have $\bigtriangleup EKO \sim \bigtriangleup EPA$ $\Rightarrow$ EK=KO $\Rightarrow$ $O \in \odot CPQE$ $\Rightarrow$ $O,P,Q,C$ are concyclic. $\Box$

$K,L$ are symmetric to $Q,P$ wrt midpoints of $BC,AC$(points $M,N$). then $CK/CL=CA/CB$ so $CKL\sim CAB$ and $KL/AB=CL/BC=PQ/AB$ so $KL=PQ$ but $OL=OP$,$OQ=OK$ so $LOK\cong POQ$ so $\angle PON=\angle POL/2=\angle QOK/2=\angle QOM$ and they have the same orientation so $\angle POQ=\angle MON=180-\angle PCQ$ so $CPOQ$ is cyclic.

Isn't it this one ? Best regards, sunken rock