It is enough to prove that the circumcenter of $PSR$ lies on the perpendicular bisector of $BC$. By dillating at $A$ with ratio $\frac{1}{2}$, we notice that this is in fact the exact same problem as IMO Shortlist 2016 G5, so use any solution from that and we're done. Extremely disappointing problem...

Interesting configuration: If we let $Q'$ be $(PRS)\cap PQ$ let $X$ be the point of intersection of the perpendicular to $BC$ and $(PRS)$ so we have $RSPX$ being an harmonic quadrilateral. Now seems like we can bash some concurrences to have $(Q,Q';(O_\perp \cap PQ),P_\infty)=-1$.

This problem is proposed by me. I found this problem playing in geogebra, moving points around the Euler line. I didn't realize it was an implication of the problem IMO Shorlist 2016 G5. Sorry

It's ok, such things happen from time to time and it's a nice problem. I would have expected the people who created the test to notice the similarity though, but those things happen too.

In my opinion this is a cool problem. I was not aware of the conection with the previous problem so I will post my solution from scratch. $\textbf{Notation:}$ Let $G,D = AH,AO\cap(ABC)$. $A'$ the reflection of $A$ onto the perpendicular bisector of $BC$. $L\in (ABC)$ such that $AL\perp OH$. $K$ such that $KB,KB$ are tangents to $(ABC)$. $T$ the second intersection point of $KG$ and $(ABC)$. $O'$ the reflection of $O$ onto $BC$. $N$ on the ray $LD$ with $D$ in between $L$ and $N$. $F=AH\cap (ARS)$ . $W$ the intersection of $OM$ with the perpendicular bisector of $RS$. $\textbf{Claim 1:}$ $GL$ bisects $OK$ which means that $OO'\cap GL$ is the center of $(BOC)$. Proof: First $AOO'H$ is a parallelogram, $\triangle HAO=\triangle HLO$, then $HOLO'$ is and isosceles trapezium which implies $D,L,O'$ are collinear and $HOLO'G$ is cyclic. We know that $AH$ and $AO$ are isogonals wrt angle $BAC$ then in $\triangle BTC$, $TG$ is the symedian and then $T,H,M,D$ are collinear. It is easy to check that $G,O,A'$ are collinear, with easy angle working we get that pencils $(DN,DA',DO,DT)$ and $(GL,GO,GA,GT)$ are equal. Now, $OO'\parallel DA'$ and $M$ is the midpoint of $OO'$ then the first one is harmonic, so $(GL,GO,GA,GT)$ is harmonic as well, then $(GK,GL,GO,GA)$ is harmonic and $OK\parallel AG$ implies that $GL$ bisects $OK$. $\textbf{Claim 2:}$ $W$ is the center of $(RPS)$. Proof: Easy angle chasing shows that there is a spiral similarty centered at $L$ which sends $(R,S)$ to $(B,C)$. The same similarity carries $P$ to $O$ and $F$ to $G$, then $HPLF$ and $POLW$ are cyclics whics means that $L,W,F$ are collinear. Now realize that $RPSLF\sim BOCLG$, and using the $claim 1$ we get that $W$ is the center of $(RPS)$. The problems is finished because $WQ=WP$ so $Q\in (RPS)$.

I found the problem very interesting. My solution is quite elementary, it involves moving points and preserving ratios, I hope it will illustrate useful ideas for training students. Denote by $G$ the reflection of $H$ onto the bisector of $BC$. Thinking in the limit case when $P=H$ (and $Q=G$), we come across the idea of defining $E$ and $F$, the points where the circle with center $H$ and radius $HA$ intersects the lines $AB$ and $AC$. We will prove the following: (1) $GE||OB$ and $GF||OC$; and then (2) When $P$ moves on $OH$, then the points $Q,R,S$ move "at the same speed" within the segments $OG$, $BE$ and $CF$, respectively, which will imply that $QR||OB$, $QS||OC$, and from this it will be easy to deduce that $\angle SQR=\angle SPR$. Proof of (1). Let $\Gamma$ be the circumcircle of the isosceles trapezoid $CBHG$. Let us prove that $E$ and $F$ belong to $\Gamma$, which is done by easy angle-chasing with directed angles mod 180: $$\measuredangle BEH=\measuredangle AEH=\measuredangle HAE=\measuredangle BCH,$$$$\measuredangle HFC=\measuredangle HFA=\measuredangle FAH=\measuredangle HBC.$$Now, by equating angles in $\Gamma$ and using the symmetry of $CBHG$ we have: $$\measuredangle AEG=\measuredangle BEG=\measuredangle BCG=\measuredangle HBC=90^o-C=\measuredangle ABO \Rightarrow GE||OB,$$$$\measuredangle GFA=\measuredangle GFC=\measuredangle GBC=\measuredangle BCH=90^o-B=\measuredangle OCA \Rightarrow GF||OC.$$Proof of (2). Let $B',C',E',F',R',S'$ be the midpoints of $AB,AC,AE,AF,AR,AS$, respectively. It is clear that $OB',PR'$ and $HE'$ are all perpendicular to $AB$, and hence they are parallel to each other. Similarly, $OC',PS'$ and $HF'$ are perpendicular to $AC$ and parallel to each other. Here comes a "chain reaction" of conservation of ratios: $$\frac{BR}{RE}=\frac{B'R'}{R'E'}=\frac{OP}{PH}=\frac{OQ}{QG} \Rightarrow QR||OB,$$$$\frac{CS}{SF}=\frac{C'S'}{S'F'}=\frac{OP}{PH}=\frac{OQ}{QG} \Rightarrow QS||OC.$$Finally, an additional calculation with central angles shows that: $$\measuredangle SQR=\measuredangle COB=2\measuredangle CAB=2\measuredangle SAR=\measuredangle SPR,$$from which it follows that $P,Q,R,S$ are concyclic.

This problem is a great complex bash exercise . Not that I have much experience with complex numbers, but it's the first time I use the concyclic formula. Solution: Let $(ABC)$ be the unit circle and the complex number for each point is it's lowercase letter. Because point $P$ lies on $OH$, we get that (from the collinearity formula) $h\bar{p} =\bar{h} p$. We also get that $q=bc\bar{p} $, from the reflection of a point over a line formula. We know that the foot of the perpendicular from $P$ to $AB$ is the midpoint of $AR$, i.e $r=b+p-ab\bar{p} $. Analogically $s=c+p-ac\bar{p} $. Now the points $P,Q,R$ and $S$ are concyclic iff $\frac{r-p}{s-p}\cdot \frac{s-q}{r-q}=\frac{\bar{r}-\bar{p}}{\bar{s}-\bar{p}}\cdot \frac{\bar{s}-\bar{q}}{\bar{r}-\bar{q}}$ . This is equivalent to $\frac{c-ac\bar{p} }{b-ab\bar{p} }\cdot \frac{b+p-ab\bar{p} -bc\bar{p} }{c+p-ac\bar{p} -bc\bar{p} }=\frac{\frac{1}{c}-\frac{1}{ac}p}{\frac{1}{b}-\frac{1}{ab}p} \cdot \frac{\frac{1}{b}+\bar{p} -\frac{1}{ab}p-\frac{1}{bc}p}{\frac{1}{c}+\bar{p} -\frac{1}{ac}p-\frac{1}{bc}p}$ Now because $\frac{c-ac\bar{p} }{b-ab\bar{p} }=\frac{c}{b}$ and $\frac{\frac{1}{c}-\frac{1}{ac}p}{\frac{1}{b}-\frac{1}{ab}p}=\frac{b}{c}$ by flipping both denominators we would want to prove that $c^2(b+p-ab\bar{p} -bc\bar{p})(\frac{1}{c}+\bar{p} -\frac{1}{ac}p-\frac{1}{bc}p)-b^2(c+p-ac\bar{p} -bc\bar{p})(\frac{1}{b}+\bar{p} -\frac{1}{ab}p-\frac{1}{bc}p)=0 \Leftrightarrow$ (by easy expanding) $(c-b)(\frac{bc}{a}p\bar{p} - abc\bar{p}^2 +ap\bar{p} - \frac{p^2}{a} - bc\bar{p}^2(b+c) + 2p\bar{p}(b+c)- \frac{1}{bc}p^2(b+c))=0 \Leftrightarrow $ $(c-b)((bc\bar{p}-p)(\frac{p}{a}-a\bar{p})-\frac{b+c}{bc}(b^2c^2\bar{p}^2- 2bcp\bar{p} + p^2))=0 \Leftrightarrow $ $(c-b)(bc\bar{p}-p)(\frac{p}{a} - a\bar{p}-\frac{b+c}{bc}(bc\bar{p}-p))=0 \Leftrightarrow $ $(c-b)(bc\bar{p}-p)(p(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-(a+b+c)\bar{p})=0$ , which is true because $p(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=(a+b+c)\bar{p}$ is equivalent to $\bar{h}p=h\bar{p}$, which we already showed holds, thus finishing the proof.$\blacksquare $

Note that $R$ is the reflection of the foot from $P$ to $AB$ across $A$. Let $f$ be the foot, so $$f=\frac{1}{2}(a+b+p-ab\overline{p}).$$This means that $r=2f-a=b+p-ab\overline{p}.$ Similarly, $s=c+p-ac\overline{p}.$ Now, the idea is the following. If we compute $q$, it does not fit well with our expressions for $p,r,s$ to apply the cyclic formula. Thus, we will try to eliminate $q$ from the problem and instead show that the circumcenter of $\triangle PRS$ lies on the perpendicular bisector of $BC$. We wish to compute the circumcenter of $p,b+p-ab\overline{p},c+p-ac\overline{p}$. Of course, subtract $p$ so we want to compute circumcenter of $$0,b(1-a\overline{p}),c(1-a\overline{p}).$$Now, divide by $1-a\overline{p}$, so we want the circumcenter of $$0,b,c.$$By the circumcenter formula for $z=0$, this is $$\frac{bc}{b+c}.$$Reversing the transformations, the circumcenter of $\triangle PRS$ is $$\frac{bc}{b+c}(1-a\overline{p})+p=\frac{bc-abc\overline{p}+pb+pc}{b+c}.$$Since the perpendicular bisector of $BC$ is the line through $0$ and $b+c$, we wish to show that $$\frac{bc-abc\overline{p}+pb+pc}{(b+c)^2}$$is real. Setting it equal to its conjugate and expanding, it becomes $$abc-a^2bc\overline{p}+apb+apc=abc-bcp+abc^2\overline{p}+ab^2c\overline{p}$$$$p(ab+ac+bc)=\overline{p}(abc)(a+b+c).$$Since $P$ lies on $OH$, we have $$\frac{p}{a+b+c}\in R$$$$\frac{p}{a+b+c}=\frac{\overline{p}}{1/a+1/b+1/c}$$$$\frac{p}{\overline{p}}=\frac{abc(a+b+c)}{ab+ac+bc}.$$This can be arranged to the above condition, so the circumcenter of $\triangle PRS$ lies on the perpendicular bisector of $BC$ and we are done.

This is an extremely straightforward complex bash. Let the perpendicular bisector of $\overline{BC}$ be the real axis. Then $p = k(a+b+c)$ for some real number $k$, and $r = b+p-ab\overline p$, $s = c+p-ac \overline p$. Furthermore, we have $q = \overline p$ by definition. So it suffices to verify that $$\frac{b-ab\overline p}{c-ac \overline p} \div \frac{b+p-ab\overline p-\overline p}{c+p-ac\overline p - \overline p} = \frac{bc-kac-\frac{kb}a - k}{bc-kab - \frac{kc}a - k}$$is a real number, which is a straightforward verification using the fact that $bc=1$.

We claim $OB\parallel QR,OC\parallel QS$. Move $P$ with degree $1$ on $HO$, so $Q,R$ have degree $1$. Thus $OB\parallel QR$ is degree $2$, so we need to check $3$ cases. If $P=O$ then $OB=QR$. If $P=\infty_{HO}$ then $QR$ is the line at infinity. If $P=H$ then $B,P,Q,C,R$ are concyclic since $\measuredangle BRC=-\measuredangle BAC=\measuredangle BPC$ so $\measuredangle BRQ=\measuredangle BCQ=\measuredangle PBC=\measuredangle ABO$, so the claim is proven. Now the result follows since $\measuredangle RQS=\measuredangle BOC$ and $\angle RPS=2\angle BAC=\angle BOC$.