Main idea is to prove that H’ is actually mid point of BC.

Seriously? This is just a combination of multiple problems put into one, giving it a false sense of high difficulty. It would have been a disaster if something like this appeared in any contest as some contestants may have a clear advantage over others. (I "solved" it while just reading the problem...) From ARMO 2009 $AF$ is a symmedian, so $P$ is the intersection of tangents to $\omega$ at $B,C$. From ARMO 2013, $H'$ is the midpoint of $BC$. Then it is a very well-known orthocenter configuration with $S$ as the HM-point, and the lines are of course concurrent on $BC$.

Just as mofumofu said this problem is just combination of many problems from All-Russian... but anyway I will use them for my solution... Let $BH$ meet $AC$ at $K$ and circle with diameter $AH$ be $ \Omega$. Claim1 : $T$ is intersection point of $ \Omega$ and $\omega$. Proof : we know $H'$ is midpoint of $BC$ and from that we know if $HH'$ meets $\omega$ at $H_1$ then $AH_1$ is diameter of $\omega$ so we have $\angle ATH = \angle 90$ so $T$ also lies on $ \Omega$. Note that for proving the concurrency we can use Radical Axis Theorem so we only need to prove $BHSC$ is cyclic but we will prove something else first. Claim2 : $BC$ is tangent to both $ASB$ and $ASC$. Proof : $H'S.H'A = H'H.H'T = H'H_1.H'T = BH'.CH' = BH'^2 = CH'^2$. Claim3 : $BHSC$ is cyclic. Proof : First Note that $K$ lies on $\Omega$. $\angle BHS = \angle 180 - \angle SHK = \angle 180 - \angle SAK = \angle 180 - \angle SAC = \angle 180 - \angle SCB$. we're Done.

Let $E'$ be the second intersection between $FD$ and $\omega$, then $E$ and $E'$ are the midpoints of minor arc $BC$ and arc $BAC$ respectively. So $(A,F;B,C) \stackrel{D}{=} (E,E';C,B) = -1$. Since $P$ lies on the $A$-symmedian of $\Delta ABC$ and $PB=PC$, $P$ is the intersection of the tangents to $\omega$ through $B$ and $C$. Let $M$ be the midpoint of $BC$, so $BMPX$ and $CMPY$ are cyclic. Note that $\angle AXM = \angle BPM = 90^{\circ} - \angle A$, so $XM \perp AY$. Similarly, $YM \perp AX$, so $M=H'$. Note that $S$ lies on the $A$-median of $\Delta ABC$, and $\angle ASH = 90^{\circ}$, so $S$ is the $A$-Humpty point. Recall that $BCHS$ is cyclic. Since $T$ lies on $HM$ and $\angle ATH = 90^{\circ}$, $T$ is the $A$-Queue point, which lies on $\omega$. Let $K$ be the foot of the altitude from $A$ to $BC$, so $HKMS$ is cyclic. By radax on $(ASHT), (BCHS), \omega$, we have that $AT,HS,BC$ are concurrent at some point $L$. Since $LB \cdot LC = LH \cdot LS = LK \cdot LM$, $(B,C;K,L)=-1$. But $(B,C;K,FQ \cap BC) \stackrel{Q}{=} (B,C;A,F) = -1$, so $L$ lies on $FQ$. $\square$

Claim: $ABFC$ is harmonic. Proof. Perform a $\sqrt{bc}$ inversion centered at $A$ followed by a reflection about the angle bisector of $\angle BAC$. Under this transformation $\triangle ABC$ remains fixed, and $F$ gets sent to the reflection of $M$ about $D$, where $M$ is the midpoint of $\overline{BC}$. Hence $\overline{AF}$ is a symmedian as desired. $\square$ Then by well known symmedian properties, $\overline{PB}$ and $\overline{PC}$ are tangent to $(ABC)$. Claim: $H'$ is the midpoint of $\overline{BC}$. Proof. Complex bash and note, $p = \frac{2bc}{b + c}$ $x = \frac{1}{2}(a + b + p - ab\overline{p})$ $y = \frac{1}{2}(a + c + p - ac\overline{p})$ However then, \begin{align*} x + y - p - \left(\frac{b+c}{2} \right) &= a + p + \frac{1}{2}(b + c)\left( 1 - a\overline{p} \right) - p - \left(\frac{b+c}{2}\right)\\ &= a - \frac{1}{2}(b + c)\left(\frac{2a}{b+c}\right)\\ &= 0 \end{align*}proving the claim. $\square$ Then $T$ is the queue point and $S$ is the humpty point. Note that $F$ is the reflection of $S$ about $\overline{BC}$ as it is well known the humpty point satisfies $\frac{SB}{SC} = \frac{AB}{AC} = \frac{FB}{FC}$. Thus it suffices to show $\overline{QF}$ and $\overline{AT}$ concur on $\overline{BC}$ but this is well known, and occurs due to harmonic chasing.

Boring problem...just recall a bunch of well-known results... It is well-known that $F$ lies on the $A$-symmedian (proof: note that $(DE)$ passes through the midpoint $M$ of $BC$ is fixed under $\sqrt{bc}$-inversion), so $P$ is the intersection of tangents to $(ABC)$ at $B, C$, so it is well-known that $M$ is actually the orthocenter of $\Delta AXY$ (proof: $\measuredangle MYA = \measuredangle MPC = 90^\circ - \measuredangle PCM = 90^\circ - \measuredangle CAB$ and similarly for the other side), so it is well-known that $AT, HS, BC$ concur (proof: they do so at the orthocenter of $\Delta AHM$) and $HS, QF$ are reflections across $BC$ (since $H,Q$ and $S,F$ are pairs of reflections across $BC$), so we are done.

By $\sqrt{bc}$ inversion, it is easy to see that $AF$ is the $A$ symmedian. This implies $P$ is the intersection of tangents at $B$ and $C$. It is well known that orthocenter of $\Delta AXY$ is midpoint of $BC$ giving $S$ to be the $A$ humpty point and $T$ to be the $A$ queue point. It is well known that $AT, HS,BC$ concur at the $A$ expoint and that $F$ is the reflection of $S$ over $BC$ and $Q$ is reflection of $H$ over $BC$ $\implies$ $QF$ is reflection of $HS$ over $BC$ hence $HS,BC,FQ$ concur and we are done.

Bruh this is like too many things spliced together as a lot of people have already commented before. Let $N_a$ be the arc midpoint of $BAC$ on $\omega$, $M_a$ the point diametrically opposite; it’s clear $M_a=E$. Now by Russia 2009 10.2 $(AF;BC)=-1$. Hence $P$ is the intersection of tangents to $\omega$ at $B$ and $C$. EGMO 2023/2 immediately dictates $H’$ midpoint of $BC$, so we have now on our hands one of the most important orthocentre configs on our hands; $T$ is the $A$-queue point and $S$ is the $A$-humpty point so actually this boring misplaced problem reduces down to showing $(SQ;BC)=-1$ which is well known.