Lemma 1: Let $P, Q$ be points in the plane of $\triangle ABC$ and denote $R \equiv BP \cap CQ$ and $S \equiv BQ \cap CP.$ Let $\ell$ be the bisector of $\angle BAC.$ If $AP, AQ$ are symmetric in $\ell$, then $AR, AS$ are symmetric in $\ell.$ Proof of Lemma: By the Dual of Desargues' Involution Theorem, there exists an involution that swaps $(AB, AC), (AP, AQ), (AR, AS).$ Since $AB, AC$ and $AP, AQ$ are symmetric in $\ell$, this involution is reflection in $\ell.$ Hence $AR, AS$ are symmetric in $\ell.$ $\blacksquare$ Under inversion at $A$, we obtain the following corollary. Corollary: Let $P, Q$ be points in the plane of $\triangle ABC$ and denote by $R$ and $S$ the second intersections of $\odot(ABP), \odot(ACQ)$ and $\odot(ABQ), \odot(ACP).$ If $AP, AQ$ are symmetric in $\ell$, then $AR, AS$ are symmetric in $\ell.$ Lemma 2: Let $P, Q$ be isogonal conjugates WRT $\triangle ABC.$ Then $\measuredangle BPC + \measuredangle BQC = \measuredangle BAC.$ Proof of Lemma: By examining $\triangle BQC$, we have \[ \measuredangle BPC + \measuredangle BQC = \measuredangle BPC + \measuredangle QBC + \measuredangle BCQ = \measuredangle BPC + \measuredangle ABP + \measuredangle PCA = \measuredangle BAC, \]where the last step follows from examining quadrilateral $ABPC.$ $\blacksquare$ _______________________________________________________________________________________________________________________________________________ Main Proof: Let $Q^*$ denote the second intersection of $\odot(BOC)$ and $\odot(AB)$ and let $R$ be the projection of $B$ onto $CA.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; 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label("$Q^*$", (1006.691773397018,1014.636722768431), NW * labelscalefactor); dot((1016.9487027379275,1016.2120584526925),linewidth(3.pt) + dotstyle); label("$Q$", (1017.0612981671627,1016.3747965225178), NE * labelscalefactor); dot((1015.4920652065816,1008.9736305721863),linewidth(3.pt) + dotstyle); label("$R$", (1015.1759300271364,1007.8317221380233), E * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Since $BR, BO$ are symmetric in the bisector of $\angle ABC$, the corollary for points $R, O$ and $\triangle ABC$ implies that $BP, BQ^*$ are symmetric in the bisector of $\angle ABC.$ Moreover, from $QA \parallel PO$, we infer that \[ \measuredangle BQ^*A + \measuredangle BQA = 90^{\circ} + \measuredangle BPO = 90^{\circ} + \measuredangle BAO = \measuredangle BCA. \]Hence, by Lemma 2, $Q^*$ is uniquely determined as the isogonal conjugate of $Q$ WRT $\triangle ABC.$ Thus, using $\measuredangle BQ^*C = \measuredangle BOC = 2\measuredangle BAC$, we obtain $\measuredangle BQC = -\measuredangle BAC$ by Lemma 2. $(\star)$ By $(\star)$ we have $\measuredangle PCQ = \measuredangle BQC - 90^{\circ} = \measuredangle OCB.$ Hence, $\measuredangle PCO = \measuredangle QCB$, as desired.

Let $\measuredangle ABP=\beta$ and $\measuredangle QAB=\theta$ $\Longrightarrow$ $\measuredangle PAQ$ $=$ $\beta+\theta$, so from $PO\parallel AQ$ and $ABPO$ is cyclic we get $\measuredangle PAQ$ $=$ $\measuredangle PQA$ $=$ $\measuredangle OAB$ $=$ $\measuredangle OBA$ $=$ $\beta+\theta...(\star)$. Let $R\equiv BP\cap \odot (ABC)$ $\Longrightarrow$ $\measuredangle ARB$ $=$ $\measuredangle ACB$ $=90^{\circ}$ $-\beta$ $-\theta$, so from $\measuredangle AQB$ $=$ $\beta$ $+$ $\theta$ we get $\measuredangle QAR$ $=$ $90^{\circ}$, but by $(\star)$ we get $PA$ $=$ $PQ$, since $\measuredangle QAR$ $=$ $90^{\circ}$ and $PA$ $=$ $PQ$ we get $PA$ $=$ $PQ$ $=$ $PR$, since $CP\perp QR$ we get $ CQ$ $=$ $CR$ $\Longrightarrow$ $\measuredangle BAC$ $=$ $\measuredangle CRQ$ $=$ $\measuredangle CQR...(\star\star)$. Since $O$ is the circumcenter of $\odot (ABC)$ we get $\measuredangle OCB$ $=$ $90^{\circ}-\measuredangle BAC$. On the other hand, so from $(\star\star)$ we get $\measuredangle PCQ$ $=$ $90^{\circ}-\measuredangle PQC$ $=$ $90^{\circ}-\measuredangle BAC$ hence $\measuredangle OCB$ $=$ $\measuredangle PCQ$ $\Longrightarrow$ $\measuredangle PCO=\measuredangle QCB.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(23.00000000000088cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 11.06000000000042, xmax = 34.06000000000130, ymin = 8.959999999999996, ymax = 20.57999999999999; /* image dimensions */ pen ttffqq = rgb(0.2000000000000002,1.000000000000000,0.000000000000000); pen ttzzqq = rgb(0.2000000000000002,0.6000000000000006,0.000000000000000); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000); draw(arc((23.90550500080287,18.35457615823758),0.4000000000000153,-136.4774096047653,-74.53634663439797)--(23.90550500080287,18.35457615823758)--cycle, ttzzqq); 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From the conditions, $A,P,O,B$ lie on a circle in this order ($\alpha>45^\circ$). Need: $\angle PCQ=\angle BCO=90^\circ-\alpha$, same as $\frac{CP}{QP}=\tan \alpha$. $(1)$ $M$ and $N$: centers of $BPC$ and $APOB$, resp. $MN$ is perpendicular to their radical axis, $BP$. Also $OM\perp BC$ and $ON\perp AB$, so $\angle ONM=\angle ABP=\varphi$, $\angle OMN=\angle CBP=\theta$. From $\triangle OMN$, we have $\frac{OM}{ON}=\frac{\sin \varphi}{\sin \theta}$. $(2)$ Let $\rho$ be the radius of circle $APOB$ and $R$ be the circumradius. $\rho=\frac{AB}{2\sin AOB}=R\cdot \frac1{2\cos \gamma}$. Since $\angle PQA=\angle QPO=\angle BPO=\angle BAO=90^\circ-\gamma$, and $\angle BPA=2\gamma$, we get $QP=AP$, so $QP=AP=2\rho \sin \varphi=R\frac1{\cos\gamma}\sin\varphi$ $(3)$ but also $CP=BC\sin \theta=2R\sin \alpha\sin \theta$, $(4)$ $\frac{OM}{ON}=\frac{R\cos\alpha}{\rho}=2\cos\alpha\cos\gamma$. $(5)$ Now $(2-5)$ imply $(1)$, so done.

Finally got around to solving this problem. Angle chasing gives $\angle AQP = 90 - \angle C$ and $\angle APB = \angle AOB = 2 \angle C$. This gives $PA=PQ$. Now if we set $R = BP \cap \omega$ ($\omega$ is the circumcircle of $\triangle ABC$) we have $\angle ARB = \angle C$, so $\triangle QAR$ is a right triangle. This gives $PQ=PR$, and it is easy to see that $CP \perp RQ$, so $\angle PCQ = 90 - \angle PRC = 90-\angle A = \angle OCB$, which gives the desired.

A bit difficult. Let $M$ be the midpoint of $BC$ and let $O'$ be the circumcenter of $\triangle BQC$. Let $S$ and $T$ be the circumcenters of $\triangle APB, \triangle AQB$ respectively. As $$\angle AQB = 180^{\circ}-\angle BPO = 90^{\circ} + \angle C$$, we get that $\angle TAB =\angle C$ or $S$ is midpoint of $OT$. But $MS$ and $O'T$ are both perpendicular to $CP$, thus $M$ is the midpoint of $OO'$. Hence $\angle BQC = 180^{\circ}-\angle A$ so $\angle PCQ=90^{\circ}-\angle A$, implying the desired isogonality.

$\textbf{Lemma}$ Define the projections of $P$ on $AB, AC$ as $X, Y$ and as $Z, W$ the projections of $Q$. Then $P, Q$ are isogonal conjugates wrt $\angle BAC$ if and only if $X, Y, Z, W$ are concyclic. Moreover the midpoint of $PQ$ is the center of this circle. The proof is a simple angle chasing exercise. The problem statement is equivalent with the isogonal conjugacy of $Q, O$ wrt $\angle PCB$. Let $M$ be the midpoint of $BC$ and $D$ be another intersection of $AQ$ with the circumcircle of $A, P, O, B$. Then $PQDO$ is a parallelogram. Denote $N$ as the intersection of its diagonals. By our lemma the statement is equivalent with showing that $ND = NP = NM$ which is equivalent to $\angle PMD = 90^{\circ}$. However, $PM= MC= MB$ and $PD = OB$ since $PODB$ is an isosceles trapezoid. By this observation also $\angle DPM = \angle OBM$, thus $\triangle OBM \equiv \triangle DPM$, hence $\angle PMD = \angle OMB = 90^{\circ }$.

Elegant conjuring problem! Posting for storage. danepale wrote: Let $ABC$ be an acute triangle such that $\angle BAC > 45^{\circ}$ with circumcenter $O$. A point $P$ is chosen inside triangle $ABC$ such that $A, P, O, B$ are concyclic and the line $BP$ is perpendicular to the line $CP$. A point $Q$ lies on the segment $BP$ such that the line $AQ$ is parallel to the line $PO$. Prove that $\angle QCB = \angle PCO$. Observe that $\angle QAO=\angle POA = \angle PBA$ and $\angle PAO=\angle PBO$ hence $\angle PAQ = \angle OBA$. Likewise, $\angle PQA = \angle QAB+\angle QBA=\angle OAB-\angle OAQ+\angle PBA=\angle OAB$ since $\angle OAQ=\angle PBA$. Thus, $PA=PQ$, and so if we reflect $Q$ across $P$, we get point $R$ such that $\angle RAQ=90^{\circ}$, and $\angle ARB=\angle ARQ = 90^{\circ}-\angle AQP=90^{\circ}-\angle OAB=\angle C$, hence $R$ lies on the circumcircle of $ABC$. Since $CP \perp QR$, it follows that $\angle CQB=180^{\circ}-\angle CQP=180^{\circ}-\angle CRP=180^{\circ}-\angle CAB$ hence $\angle QCP=(180^{\circ}-\angle CAB)-90^{\circ} = 90^{\circ}-\angle CAB=\angle OCB$ so $CO, CQ$ are isogonal in angle $PCB$, which proves our claim.