Work in the complex plane. Set $\odot ABC$ as the unit circle, and let $a^2,b^2,c^2$ be the affixes of $A,B,C$. Let lowercase letters denote the complex affixes for the other points. Suppose without loss of generality the midpoint $M$ of arc $BAC$ has affix $bc$. Since $Q=AM \cap BC$, we find $q=\frac{a^2b^2+a^2c^2-a^2bc-b^2c^2}{a^2-bc}$. From $\triangle BPA \sim \triangle APC$, $\frac{p-a^2}{p-b^2}=\frac{p-c^2}{p-a^2} \Longrightarrow p=\frac{a^4-b^2c^2}{2a^2-b^2-c^2}$. We compute \[p-a^2=-\frac{(a^2-b^2)(a^2-c^2)}{2a^2-b^2-c^2}\] \[q-b^2=\frac{c(a^2-b^2)(c-b)}{a^2-bc}\] \[q-p=\frac{(a^2-b^2)(a^2-c^2)(b^2+c^2-bc-a^2)}{(2a^2-b^2-c^2)(a^2-bc)}\] Now, $\angle QPA + \angle OQB = 90^{\circ} \Longleftrightarrow \frac{q-p}{p-a^2} \cdot \frac{o-q}{q-b^2} \in i \mathbb{R}$. From the computations above, \[\frac{q-p}{p-a^2} \cdot \frac{o-q}{q-b^2}=\frac{(b^2+c^2-bc-a^2)(a^2b^2+a^2c^2-a^2bc-b^2c^2)}{c(a^2-bc)(c-b)(a^2-b^2)} \] The last expression changes sign under conjugation, implying it is pure imaginary as desired.

Let $ M $ and $N $ be the midpoints of $ BC $ and its minor arc respectively . Then $ A,Q,N,M $ are concyclic . Note also that $ AP $ is $ A$-symmedian of $ ABC $ . $ \angle PAQ = 180^o - \angle MAQ = \angle MNQ $ . It is therefore sufficient to prove that $ \Delta QAP $ ~ $ \Delta QNO $ . Consider $ QA / QN = \cos(\angle AMO) $ , on the other hand , $ \angle AMO = \angle OAP $ . Extend $ AP $ and it meets the circumcircle again at $ P' $ then $ \Delta CPP' $ ~ $ \Delta CAB $ and $ \Delta BCP' $ ~ $ \Delta BAP $ , $ AP:AB = CP':CB = PP':AB ~ \implies AP=PP' $ so $ OP \bot AP $ and $ AP/NO = \cos(OAP) = QA/QN $ and we are done .

Another solution can be obtained by noting that P is the midpoint of the common chord of the A-Apollonius circle and the circumcircle of triangle ABC. After this, let X be the circumcentre of the A-Apollonius circle and U be the second intersection of this circle with the circumcircle of ABC. Then X is the circumcentre of AUQ and X lies on BC. P is the midpoint of side AU in this triangle. Work in the frame of reference of triangle AUQ, and it's not difficult to finish from here.

Let $\triangle M_AM_BM_C$ be the medial triangle of $\triangle ABC$ and let $\omega$ be the circle of diameter $\overline{AO}$ passing through $M_B, M_C.$ Let $X, T, K$ be the second intersections of $AQ, AM_A, XM_A$ with $\omega.$ Since $\triangle PAB \cup M_C \sim \triangle PCA \cup M_B$, we obtain $\measuredangle PM_CA = \measuredangle PM_BC \implies P \in \omega.$ Meanwhile, we also obtain $\text{dist}(P, AB) : \text{dist}(P, AC) = AM_C : AM_B.$ Therefore, it is well-known (Characterization 2) that $P$ lies on the $A$-symmedian in $\triangle AM_BM_C.$ Then since $T$ lies on the $A$-median in $\triangle AM_BM_C$, it follows that $AP$ and $AT$ are isogonal WRT $\angle M_BAM_C.$ Therefore, $PT \parallel M_BM_C.$ Finally, note that $X$ is the midpoint of arc $\widehat{M_BAM_C}$ on $\omega$ because $AX$ is the external bisector of $\angle M_BAM_C.$ Hence, if $Q' \equiv AX \cap PK$, Pascal's Theorem on cyclic hexagon $AXXKPT$ yields $Q'M_A \parallel M_BM_C.$ Therefore, $Q' \equiv Q.$ Thus, $P, K, Q$ are collinear, and it follows that $\measuredangle APQ = \measuredangle APK = \measuredangle AXK.$ Meanwhile, as $\omega$ is the circle of diameter $\overline{AO}$, we have $\angle OXQ = \angle OM_AQ = 90^{\circ}.$ Therefore, $O, Q, X, M_A$ are concyclic. Hence, $\measuredangle M_AQO = \measuredangle M_AXO = \measuredangle KXO.$ Thus, \[\measuredangle APQ + \measuredangle M_AQO = \measuredangle AXK + \measuredangle KXO = \measuredangle AXO = 90^{\circ}. \; \square\]

Let $w$ be the circumcircle of triangle $ABC$ the tangants from $B,C$ intersect eachother at point $M$ let the line $AM$ intersect $w$ at point $D$ . At first proof that $P$ is unic and it is the midpoint of $AD$ . Let $T,T'$ be the midponts of arcs $.BAC,BC$ . Lines $TA,T'D$ intersect each other at point $Q$ . and $O$ is the midpoint of segment $TT'$ . Triangles $QAD,TQT'$ are similar and points $P,O$ are the midpoints of $AD,TT'$ so angles $QPA,QOT'$ are equal . We know that$OT'$ is perpendicular to $BC$ so $QOT'+OQB=90$ so we find that $OQB+QPA=90$

Let the $A$-symmedian intersect $\Gamma=(ABC)$ at $D$. It is well-known that the $A$-Apollonius circle $ADQ$ (call it $\omega$) is orthogonal to $\Gamma$ and that $P$ is the midpoint of $AD$, implying $O$ is the inverse of $P$ with respect to $\omega$. Let $S=AA\cap BC$ be the center of $\omega$. Then \[ \angle OQB=\angle QPS=90-\angle APQ \]as desired.

Invert about $A$ and denote the image of $X$ as $X'$ for all points $X$. The given condition about $P$ implies that the circumcircle of $BPA$ is tangent to $AC$ at $A$ and the circumcircle of $APC$ is tangent to $AB$ at $A$. These circles will invert to lines parallel to $AC$ and $AB$ going through $B'$ and $C'$ respectively making $AB'P'C'$ a parallelogram. Since $Q$ was on $BC$, $Q'$ is on the circumcircle of $AB'C'$ and is in fact diametrically opposite the midpoint of minor arc $B'C'$ (from angle chasing). Finally extend $Q'A$ to meet $B'C'$ at $R$. If we let $O''$ denote the circumcenter of $AB'C'$ (this is not the image of $O$), then it suffices to show by simple properties of inversion that $$O''RB' + AQ'P' = 90$$Reflect $P'$ over $Q'O''$ to get $P''$. Then it suffices to show $$O''Q'P'' = O''RQ'$$To do this, observe that $P''$ is also the reflection of $A$ across $BC$ which yields $AQ'O'' ~ AP''R$ making $A$ the center of spiral similarty mapping $Q'O''$ to $P''R$. Hence by the spiral similarity lemma, if $X$ denotes the intersection of $Q'P''$ and $O''R$ then we have $AXO''Q'$ and $AXP''R$ are both cyclic. Thus $$O''Q'P'' = XP''A = O''RQ'$$as desired.

Let $L=AA \cap BC$ and let $D$ be the feet of the $\angle A $ bisector on $BC$. Then $(QDA)$ is the $A$ apollonius circle(let its be $\omega$) and $L$ is the center of $\omega$. Let it intersect $\odot (ABC)$ at $K$. Then $AK$ is the $A$ symmedian in $\triangle ABC$. From the given angle condition it is clear that $P$ is the midpoint of $AK$. Since $\omega$ and $\odot(ABC)$ are orthogonal , inversion around $L $ swaps ${O,P}$. Hence $\angle BQO=\angle LQO=\angle LPQ=90-\angle QPA$. $\square$.

ELMOSL 2012 G7 wrote: Let $\triangle ABC$ be an acute triangle with circumcenter $O$ such that $AB<AC$, let $Q$ be the intersection of the external bisector of $\angle A$ with $BC$, and let $P$ be a point in the interior of $\triangle ABC$ such that $\triangle BPA$ is similar to $\triangle APC$. Show that $\angle QPA + \angle OQB = 90^{\circ}$. Alex Zhu. Notice that $P$ is the $A-\text{Dumpty Point}$ of $\triangle ABC$. So, $OP\perp BC$ and $P\in A-\text{Symmedian}$. So, it just suffices to show that $\angle OQB=\angle QPX\implies \angle QOP=\angle PQC$. So we just have to show that $\odot(OPQ)$ is tangent to $BC$ at $Q$. Let $OP\cap BC=X$ and $AY$ be the bisector of $\angle BAC$ where $Y\in BC$. So, $X$ is the Circumcenter of the $A-\text{Appolonius Circle.}$ Also $P\in\odot(BOC)$ and $(QY;BC)$ is harmonic. So, Combining MC'laurin and PoP we get that $$XP\cdot XO=XB\cdot XC=XQ^2\implies \odot(OPQ)\text{ is tangent to } BC \text{at Q.}\blacksquare$$

$\sqrt{bc}$ inversion and reflect across the internal bisector of $A$. $O'$ is the reflection of $A$ over $BC$ , $\square ABP'C$ is a parallelogram. $Q'$ is the intersection of the external bisector of $\angle A$ and the circumcircle. $R$ is the reflection of $Q'$ over $BC$. \[\angle QPA + \angle OQB = 90^{\circ} \iff \angle AQ'P' + \angle AB'Q' - \angle AO'Q' =90^{\circ} \iff \angle AO'Q' = \angle RQ'P'\]which equal to $\angle ARQ'$.

Let the $AO$ meet $(ABC)$ again at $A_1$, the foot of the $A$-altitude be $D$, the midpoint of $BC$ be $M$, and $X = AO \cap BC$. Clearly, $P$ is the $A$-Dumpty point. Now, we consider $\sqrt{bc}$-inversion. Notice that $O^*$ is the reflection of $A$ over $BC$, $P^*$ is the reflection of $A$ over $M$, $Q^*$ is the midpoint of arc $BAC$, and $X^*$ is the second intersection between $AD$ and $(ABC)$. Because $DM \parallel O^*P^*$ from midlines, $$\angle AO^*P^* = \angle ADM = 90^{\circ}$$so $M$ is the circumcenter of $(AO^*P^*)$, which implies $MO^* = MP^*$. Furthermore, we have $$OM \perp BC \parallel O^*P^*$$which means $OM$ is the perpendicular bisector of $O^*P^*$. It's easy to see that $X^*$ and $A_1$ are also symmetric about $OM$. Now, since $Q^*$ lies on $OM$, we have $$\angle QPA = \angle AQ^*P^* = \angle AQ^*A_1 - \angle P^*Q^*A_1 = 90^{\circ} - \angle O^*Q^*X^*$$$$= 90^{\circ} - (\angle AQ^*O^* - \angle AQ^*X^*) = 90^{\circ} - (\angle AOQ - \angle AXQ)$$$$= 90^{\circ} - \angle OQX = 90^{\circ} - \angle OQB$$as desired. $\blacksquare$ Remark: I'm still quite amazed that I managed to solve this without paper!

I absolutely despise this problem, and I don't know why It is clear that $P$ is the $A-$Dumpty point of $\triangle ABC$. Let $OP \cap BC=X$ and $AP\cap (ABC)=Y$ and let the $A-$angle bisector intersect $BC$ at $D$ Note that since $(A,Y;B,C)=-1$ we get that $X$ is just the intersection of the tangents from $A$ and $Y$ to $(ABC)$ as $P$ is the midpoint of $AY$ Now it is well known that the centre of the $A-$Appolonius circle is $X$ (I actually didnt know that and had to prove it but I'm lazy now xD) Let $OQ\cap(AQYP)=Z$, it is clear that $QZ$ is the $Q-$Symmedian in $\triangle QAY$ because $(AQYP)$ is orthogonal to $(ABC)$ This means that $$\angle QPA+\angle OQB=\angle QPA+\angle ZQY-\angle DEY=\angle QPA+\angle PQA-\angle DAJ=180^{\circ}-\angle YAQ-\angle DAY=180^{\circ}-\angle DAQ=90^{\circ}$$ This is my most garbage looking solution yet, and I think this problem deserves that

Notice that $P$ is the $A-Dumpty$ point so if we let $K$ be the intersection of the $A-symeddian$ with $(ABC)$ then $P$ is the midpoint of $AK$. (We cite this as well-known but it is easy as $\sqrt{bc}$ inversion maps $P$ with some point $A'$ satisfying that $ACA'B$ is a parallelogram, which concludes the proof of the claim). Now let $D$ be the feet of the interior bisector of $\angle BAC$ and $E$ be the midpoint of $QD$. See that $(Q,D;B,C) = -1 \implies ED^2 = EB\cdot EC$. But now, as $\angle QAD = 90^\circ$ this implies that $E$ is the center of $(QAD)$. Thus, $EA$ is tangent to $(ABC)$. Now, by LaHire, as $(A,K;B,C) = -1$ we also get that $EK$ is tangent to $(ABC)$. Thus, we get $E- P - O$, while at the same time $EP\cdot EO = EQ^2 \implies \angle QPE = \angle BOQ$, so we are done as $\angle APQ = 90^\circ - \angle QPE$ ends the problem.

I present a solution without inversion and projective geometry and with the introduction of only one point(apart from proving well known lemmas), so this is probably the easiest solution to find for beginners on this thread

Note that $P$ is $A-$dumpty. After performing $\sqrt{bc}$ inversion and reflecting over the angle bisector of $\measuredangle CAB$, $Q^*$ lies on the perpendicular bisectors of $BC,O^*P^*$ thus, $(OPQ)$ and $BC$ are tangent to each other where $BCP^*O^*$ is an isosceles trapezoid. \[\measuredangle OQB+\measuredangle QPA=180-\measuredangle QPO+\measuredangle QPA=90\]As desired.$\blacksquare$