Cute problem! Denote $\Gamma_9$ by the nine-point circle of $\bigtriangleup ABC$ and $\text{Pow}(P, \mathcal{C})$ by the power of a point $P$ w.r.t. circle $\mathcal{C}$. Claim 01 $\odot(HQP)$ and $\Gamma_9$ are tangent at $H$. Proof. Note that $P,Q,M,N$ are concyclic because $\angle QPM = \angle QCB = \angle GNM$. Consider three circles $\odot(ABC), \odot(AMN), \odot(PQMN)$. The pairwise radical axes of these circles are $PQ, MN$ and the tangent of $\odot(ABC)$ at $A$. Let their radical center be $D = PQ \cap MN$. Since $A$ is reflection of $H$ w.r.t. $MN$, therefore $DH^2 = DA^2 = DP\times DQ = DM \times DN$. Therefore, $DH$ are both tangent to $\odot(PQH)$ and $\odot(MNH) \equiv \Gamma_9$, so the circles are tangent at $H$ as well. $\square$ Claim 02 $\odot(HNQ), \odot(HMP), \odot(ABC)$ are concurrent on $GH$. Consider the inversion with center $G$ and radius $\sqrt{\frac{\text{Pow}(G, \odot(ABC))}{2}}=\sqrt{GN\times GQ}$. This inversion maps $\Gamma_9$ to $\odot(ABC)$. Let $H$ gets inverted to $I$. Since $H \in \Gamma_9$, so $I \in \odot(ABC)$. Therefore we have: $GN \times GQ = GM \times GP = GH \times GI \implies \odot(HNQ), \odot(HMP), \odot(ABC)$ are concurrent at $I$. $\square$ Let $E, F$ be the intersection point of $\Gamma_9$ with $HQ, HP$ respectively. Since $\odot(QNH), \odot(HNG)$ and $\Gamma_9$ are coaxial, therefore: \begin{align*} \frac{\text{Pow}(Q, \odot(HNG))}{\text{Pow}(Q, \Gamma_9)} &= \frac{\text{Pow}(I, \odot(HNG))}{\text{Pow}(I, \Gamma_9)} \\ \implies \frac{QS \times QH}{QE \times QH} &= \frac{IH \times IG}{\text{Pow}(I, \Gamma_9)} \\ \implies \frac{QS}{QE} &= \frac{IH \times IG}{\text{Pow}(I, \Gamma_9)} \end{align*}Similarly, we get: $\frac{PR}{PF} = \frac{IH \times IG}{\text{Pow}(I, \Gamma_9)}$. Therefore, $\frac{PR}{PF} = \frac{QS}{QE}$. By Claim 01, we infer $EF || PQ$. This means $RS || PQ$ as well, as desired. $\blacksquare$ [asy][asy] size(12cm); defaultpen(fontsize(9pt)); pair A = dir(110), B = dir(200), C = dir(-20); pair H = foot(A, B, C), G = (A+B+C)/3, N = (A+B)/2, M = (A+C)/2; pair P = intersectionpoints(circumcircle(A, B, C), B--(B+10*(M-B)))[0]; pair Q = intersectionpoints(circumcircle(A, B, C), C--(C+10*(N-C)))[0]; pair I = intersectionpoints(circumcircle(A, B, C), G--(G+10*(H-G)))[0]; pair S = intersectionpoints(circumcircle(N, G, H), Q--H)[0]; pair R = intersectionpoints(circumcircle(M, G, H), P--H)[0]; pair E = intersectionpoints(circumcircle(H, N, M), Q--H)[0]; pair F = intersectionpoints(circumcircle(H, N, M), P--H)[0]; draw(circumcircle(A, B, C), heavymagenta); draw(circumcircle(N, G, H), green); draw(circumcircle(M, G, H), green); draw(circumcircle(H, N, M), blue); draw(circumcircle(H, Q, P), lightblue); draw(circumcircle(Q, N, H), red); draw(A--B--C--A^^C--Q^^B--P^^G--I^^Q--H^^P--H, black); string[] names = {"$A$", "$B$", "$C$", "$H$", "$M$", "$N$", "$G$", "$Q$", "$P$", "$I$", "$S$", "$E$", "$F$", "$R$"}; pair[] points = {A, B, C, H, M, N, G, Q, P, I, S, E, F, R}; pair[] ll = {A, B, C, H, M, N, G, Q, P, I, S, E, F, R}; int pt = names.length; for (int i=0; i<pt; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy]

Thank you for your interest, and its a nice solution! I will withhold my proof a bit longer for others to try first!

Seicchi28 wrote: Cute problem! Denote $\Gamma_9$ by the nine-point circle of $\bigtriangleup ABC$ and $\text{Pow}(P, \mathcal{C})$ by the power of a point $P$ w.r.t. circle $\mathcal{C}$. Claim 01 $\odot(HQP)$ and $\Gamma_9$ are tangent at $H$. Proof. Note that $P,Q,M,N$ are concyclic because $\angle QPM = \angle QCB = \angle GNM$. Consider three circles $\odot(ABC), \odot(AMN), \odot(PQMN)$. The pairwise radical axes of these circles are $PQ, MN$ and the tangent of $\odot(ABC)$ at $A$. Let their radical center be $D = PQ \cap MN$. Since $A$ is reflection of $H$ w.r.t. $MN$, therefore $DH^2 = DA^2 = DP\times DQ = DM \times DN$. Therefore, $DH$ are both tangent to $\odot(PQH)$ and $\odot(MNH) \equiv \Gamma_9$, so the circles are tangent at $H$ as well. $\square$ Claim 02 $\odot(HNQ), \odot(HMP), \odot(ABC)$ are concurrent on $GH$. Consider the inversion with center $G$ and radius $\sqrt{\frac{\text{Pow}(G, \odot(ABC))}{2}}=\sqrt{GN\times GQ}$. This inversion maps $\Gamma_9$ to $\odot(ABC)$. Let $H$ gets inverted to $I$. Since $H \in \Gamma_9$, so $I \in \odot(ABC)$. Therefore we have: $GN \times GQ = GM \times GP = GH \times GI \implies \odot(HNQ), \odot(HMP), \odot(ABC)$ are concurrent at $I$. $\square$ Let $E, F$ be the intersection point of $\Gamma_9$ with $HQ, HP$ respectively. Since $\odot(QNH), \odot(HNG)$ and $\Gamma_9$ are coaxial, therefore: \begin{align*} \frac{\text{Pow}(Q, \odot(HNG))}{\text{Pow}(Q, \Gamma_9)} &= \frac{\text{Pow}(I, \odot(HNG))}{\text{Pow}(I, \Gamma_9)} \\ \implies \frac{QS \times QH}{QE \times QH} &= \frac{IH \times IG}{\text{Pow}(I, \Gamma_9)} \\ \implies \frac{QS}{QE} &= \frac{IH \times IG}{\text{Pow}(I, \Gamma_9)} \end{align*}Similarly, we get: $\frac{PR}{PF} = \frac{IH \times IG}{\text{Pow}(I, \Gamma_9)}$. Therefore, $\frac{PR}{PF} = \frac{QS}{QE}$. By Claim 01, we infer $EF || PQ$. This means $RS || PQ$ as well, as desired. $\blacksquare$ [asy][asy] size(12cm); defaultpen(fontsize(9pt)); pair A = dir(110), B = dir(200), C = dir(-20); pair H = foot(A, B, C), G = (A+B+C)/3, N = (A+B)/2, M = (A+C)/2; pair P = intersectionpoints(circumcircle(A, B, C), B--(B+10*(M-B)))[0]; pair Q = intersectionpoints(circumcircle(A, B, C), C--(C+10*(N-C)))[0]; pair I = intersectionpoints(circumcircle(A, B, C), G--(G+10*(H-G)))[0]; pair S = intersectionpoints(circumcircle(N, G, H), Q--H)[0]; pair R = intersectionpoints(circumcircle(M, G, H), P--H)[0]; pair E = intersectionpoints(circumcircle(H, N, M), Q--H)[0]; pair F = intersectionpoints(circumcircle(H, N, M), P--H)[0]; draw(circumcircle(A, B, C), heavymagenta); draw(circumcircle(N, G, H), green); draw(circumcircle(M, G, H), green); draw(circumcircle(H, N, M), blue); draw(circumcircle(H, Q, P), lightblue); draw(circumcircle(Q, N, H), red); draw(A--B--C--A^^C--Q^^B--P^^G--I^^Q--H^^P--H, black); string[] names = {"$A$", "$B$", "$C$", "$H$", "$M$", "$N$", "$G$", "$Q$", "$P$", "$I$", "$S$", "$E$", "$F$", "$R$"}; pair[] points = {A, B, C, H, M, N, G, Q, P, I, S, E, F, R}; pair[] ll = {A, B, C, H, M, N, G, Q, P, I, S, E, F, R}; int pt = names.length; for (int i=0; i<pt; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] nice job!

Let $W$ so-called $A$-Why point of $\triangle ABC$, note that $W$ lies on $\odot(ABC)$ and $\overline{HG}$. Let $A'$ be the second intersection of $\odot(ABC)$ and $\overline{HG}$. Also note that $\overline{AA'}\parallel \overline{BC}$. Thus, by simple PoP

, $QNHW$ and $PMHW$ are cyclic. Thus, by spiral sim, $\triangle QNW\sim\triangle SNG\implies \triangle QNS\sim\triangle WNG$ and $\triangle PMW\sim\triangle RMG\implies \triangle PMR\sim\triangle WMG$. Claim. [$\diamondsuit$] $QN\cdot WM^2\cdot PG=PM\cdot WN^2\cdot QG$. Proof. By DDIT on complete quadrilateral $QNMBCG$, we get that $\angle NWB=\angle CWM$ as $\overline{AA'}\parallel \overline{BC}$. Thus, \begin{align*} \frac{WM^2}{WN^2}=\frac{\tfrac{\sin^2{\angle WCM}\cdot CM^2}{\sin^2{\angle CWM}}}{\tfrac{\sin^2{\angle WBN}\cdot BN^2}{\sin^2{\angle BWN}}}=\frac{MA\cdot MC}{NA\cdot NB}=\frac{MP\cdot MB}{NQ\cdot NC}=\frac{PM}{QN}\cdot \frac{BG}{CG}=\frac{PM}{QN}\cdot \frac{QG}{PG}. \end{align*} Hence, \begin{align*} \frac{QS}{QH}&=\frac{QS^2}{QN\cdot QG}\\&=\frac{GW^2}{WN^2} \cdot \frac{QN}{QG}\\&\overset{\diamondsuit}{=}\frac{GW^2}{WM^2} \cdot \frac{PM}{PG}\\&=\frac{PR^2}{PM\cdot PG}\\&=\frac{PR}{PH}, \end{align*}we conclude that $\overline{PQ}\parallel \overline{RS}$. $\blacksquare$

I have a solution by inversion Apply an inversion about $H$ with radius $\sqrt{-HB\cdot HC}$ that swaps $B$ and $C$ and preserves $\omega$, then we get the following problem: ''Let $\omega$ be the circumcircle of an actue triangle $ABC$. Let $H$ be the feet of aliitude from $A$ to $BC$. Let $M$ and $N$ be the reflection of $H$ about $AC$ and $AB$. The circles $(BHM)$ and $(CHN)$ intersect each other at $G$ and intersect $\omega$ at $P$ and $Q$ respectively. The lines $MG$ and $NG$ intersect the segments $HP$ and $HQ$ again at $R$ and $S$ respectively. Prove that $PQ\parallel RS$.'' \textit{Claim:} $PQMN$ is an isosceles trapezoid with $PQ\parallel MN$. \textit{Proof:} Denote $X$ and $Y$ be the projection of $H$ to $AC$ and $AB$, then since $AHCM$ is cyclic, then apply Radical Axis Theorem on the circles $(AHCM), (APCB), (MPHB)$, we get $B, X, P$ are colinear. Likewise $C, Y, Q$ are colinear. Furthermore, note that $BXYC$ is cyclic since $\angle AYX=\angle AHX=\angle ACB$, so angle subtended by arcs $AP$ and $AQ$ are equal, hence $AP=AQ$. Observe that $AXPM$ is cyclic since $\angle PAX=\angle PBH=\angle XMP$, so $\angle QPM=\angle APQ+\angle APM=\angle APQ+\angle AXM=90^{\circ}+\angle APQ$. Likewise, $\angle PQN=90^{\circ}+\angle AQP$, so since $AP=AQ$ then $\angle QPM=\angle PQN$. Then since $AM=AH=AN$ as well, we conclude that $PQMN$ is an isosceles trapezoid with $PQ\parallel MN$. $\square$ Now consider the point $L=PM\cap QN$, then $L$ is the radical center of $(BHPM), (CHQN)$ and $(PQMN)$, so $L=PM\cap QN, H=PR\cap QS, G=MR\cap NS$ are colinear. Hence by Desargues Theorem, the triangles $\triangle MPR$ and $\triangle NQS$ are perspective, so $MN\parallel PQ\parallel PS$ must hold. This completes the proof. $\blacksquare$.

Similar to above but without angle chasing. Claim: $(HPQ)$ and $(HMN)$ are tangent to each other. Proof: Invert with $\frac{\sqrt{bc}}{2}$ and reflect over the angle bisector of $\measuredangle A$. $(HMN)$ swaps with $(BOC)$ and $(HPQ)$ swaps with $(OP^*Q^*)$ where $P^*=(ABM)\cap MN, \ Q^*=(ACN)\cap MN$. $P^*B,Q^*C$ and the tangents to $(ABC)$ at $B,C$ are concurrent. Since circumcenters of $(OP^*Q^*),(OBC)$ lie on the perpendicular bisector of $BC$ where $O$ lies on, we get that those circles are tangent.$\square$ Claim: $M,N,P,Q$ are concyclic. Proof: $\frac{GM.GP}{GN.GQ}=\frac{GB}{GC}.\frac{GP}{GQ}=1\implies GM.GP=GN.GQ$ which yields the desired result.$\square$ Now take the inversion centered at $H$ with any radius. New Problem Statement: $H$ is the altitude from $A$ to $BC$ on $\triangle ABC$. Reflections of $H$ with respect to $AB,AC$ are $N,M$. $(ABC)$ intersects $(BHM),(CHN)$ at $P,Q$ and $(BHM)$ meets $(CHN)$ at $G$. If $GM,GN$ intersect $HP,HQ$ at $R,S$, then prove that $PQ\parallel RS$. Proof: By first claim, we conclude that $MN\parallel PQ$. By second claim, we see that $M,N,P,Q$ are concyclic in the inverted configuration. Apply Desargues theorem on $\triangle PRM$ and $\triangle QSN$. Since radical axises of $(MNPQ),(HGNQ),(HGMP)$ are concurrent, we get that $PR\cap QS=H, \ RM\cap SN=G, \ PM\cap QN$ are collinear hence these triangles are perspective. This implies that $PQ,SR,MN$ are concurrent. Since $MN\parallel PQ,$ we have $PQ\parallel SR$ as desired.$\blacksquare$