This problem was proposed by Burii.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -9.31, xmax = 13.35, ymin = -11.43, ymax = 4.03; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((-2.385343454044559,-1.7670989477335564),0.6,-14.642011078153248,71.26331279005197)--(-2.385343454044559,-1.7670989477335564)--cycle, linewidth(1) + qqwuqq); draw(arc((3.3180087219923147,-1.0398676072859723),0.6,132.6465081723639,218.5518320405691)--(3.3180087219923147,-1.0398676072859723)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw(circle((1.35,-0.61), 4.392083787907512), linewidth(1)); draw(circle((1.032472685424509,-5.843468072988077), 5.319610174490567), linewidth(1)); draw((-2.385343454044559,-1.7670989477335564)--(3.3180087219923147,-1.0398676072859723), linewidth(1)); draw((0.46633263397387786,-1.403483277509764)--(1.0324726854245092,-5.843468072988077), linewidth(1)); draw((1.222604575681927,-2.7097245125344767)--(1.0324726854245092,-5.843468072988077), linewidth(1)); draw((-2.385343454044559,-1.7670989477335564)--(6.007423545326369,-3.959823639952665), linewidth(1)); draw((-4.1600629244443885,-6.999247614590621)--(3.3180087219923147,-1.0398676072859723), linewidth(1)); draw((-3.0638993343228615,-2.4496518310011304)--(5.509108485686716,-2.969797194067822), linewidth(1)); draw((-0.67,3.29)--(-4.1600629244443885,-6.999247614590621), linewidth(1)); draw((-0.67,3.29)--(6.007423545326369,-3.959823639952665), linewidth(1)); /* dots and labels */ dot((-0.67,3.29),dotstyle); label("$A$", (-0.59,3.49), NE * labelscalefactor); dot((-2.6258765062780887,-2.476227747844817),dotstyle); label("$B$", (-2.43,-2.39), NE * labelscalefactor); dot((5.071085657641943,-2.943221277224136),dotstyle); label("$C$", (4.47,-2.83), NE * labelscalefactor); dot((-2.385343454044559,-1.7670989477335564),dotstyle); label("$P$", (-2.75,-1.59), NE * labelscalefactor); dot((1.222604575681927,-2.7097245125344767),linewidth(4pt) + dotstyle); label("$M$", (1.31,-2.55), NE * labelscalefactor); dot((3.3180087219923147,-1.0398676072859723),linewidth(4pt) + dotstyle); label("$Q$", (3.39,-0.87), NE * labelscalefactor); dot((1.0324726854245092,-5.843468072988077),linewidth(4pt) + dotstyle); label("$S$", (0.97,-6.29), NE * labelscalefactor); dot((-4.1600629244443885,-6.999247614590621),linewidth(4pt) + dotstyle); label("$D$", (-3.93,-7.29), NE * labelscalefactor); dot((6.007423545326369,-3.959823639952665),linewidth(4pt) + dotstyle); label("$E$", (6.09,-3.79), NE * labelscalefactor); dot((-3.0638993343228615,-2.4496518310011304),linewidth(4pt) + dotstyle); label("$B_1$", (-3.69,-2.35), NE * labelscalefactor); dot((5.509108485686716,-2.969797194067822),linewidth(4pt) + dotstyle); label("$C_1$", (5.59,-2.81), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $E = \overline{PM} \cap\overline{AC}$ and $D = \overline{QM} \cap \overline{AB}$. The condition $AP > PB$ implies $C$ is between $A$ and $E$, and $B$ is also between $A$ and $D$ as well. Since $\angle DPE = 180^{\circ} - \angle APM = 180^{\circ} - \angle AQM = \angle DQE$, the quadrilateral $DPQE$ is cyclic. Let $\overline{BC}$ cut $\odot(DPQE)$ at two points $B_1$ and $C_1$. By the converse of Butterfly Theorem, $M$ is also the midpoint of $\overline{B_1C_1}$. Therefore, $S$ is the intersection of perpendicular bisectors of $\overline{PQ}$ and $\overline{B_1C_1}$, i.e. $S$ is the center of $\odot(DPQE)$. Therefore, we conclude that $$ \angle BAC + \angle QSP = \angle BAC + \angle QDP + \angle QEP = \angle MQE + \angle QEM = \angle QMP,$$as desired. $\blacksquare$

Let $AC\cap PM=D, AB\cap QM=E$.$\angle EPD=180-\angle APM=180-\angle AQM=\angle EQD \implies PQED$ is cyclic. Let $(PQED)\cap BC=G,H$. Since $M$ is midpoint of $BC$, from Butterfly Theorem $M$ is midpoint of $GH \implies S$ is center of $(PQDE)$. So $\angle BAC + \angle QSP=(\angle BAC+\angle PEQ)+\angle PDQ=\angle EQD+\angle PDQ=\angle EMD = \angle QMP$,as desired.

Let $QM\cap AP=T,PM\cap AQ=R.$ By the converse of Butterfly theorem $\odot (PQRT)$ intersects $BC$ by two points, symmetric wrt $M,$ so $S$ is the center of this circle and $$\angle BAC+\angle QSP=\angle BAC+\angle ARM+\angle ATM=\angle RMT=\angle QMP$$

Cute problem! Let $U= MP \cap AC, V= MQ \cap AB$. Since $\angle MPA= \angle AQM \implies PQUV$ is cyclic. Let $BC$ intersect the circumcircle of $(PQUV)$ at $Z,W$. By the converse of the Butterfly Theorem, we have that $M$ is also the midpoint of $ZW$. Hence, the perpendicular bisector of $BC$ coincides with the perpendicular bisector of $ZW$. Thus, $S$ is the circumcenter of $PQUV$. Therefore, $\angle BAC+ \angle QSP= \angle MAB+ \angle MAC+ \angle PUQ+ \angle PVQ= \angle AMQ+ \angle AMP= \angle QMP$, as desired. $\blacksquare$

Let $PM$ meet $AC$ at $E$ and $QM$ meet $AB$ at $D$. Note that $\angle MPA = \angle AQM$ means $DPQE$ is cyclic. Let $BC$ meet $DPQE$ at $F,G$. Claim $: FB = GC$. Proof $:$ Note that $FB.BG = PB.BD$ and $GC.CF = QC.CE$ so we need to prove $\frac{PB}{QC} = \frac{CE}{BD}$. Note that $\frac{PB}{QC} = \frac{\sin{PMB}}{\sin{QMC}} = \frac{\sin{CME}}{\sin{QMC}} = \frac{CE}{BD}$. Now that $FB = GC$ we have $FM = MG$ so $S$ lies on perpendicular bisector of $PQ$ and $FG$ so $S$ is center of $DPQE$. Now we have $\angle QSP = \angle MDP + \angle MEQ$ so $\angle BAC + \angle QSP = \angle MDP + \angle MEQ + \angle BAC = \angle DME = \angle QMP$.

The configuration was similar to IMO Shortlist 2018/G2.

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... Let $Y=QM\cap AB$. Let $X=PM\cap AC$. Let $Z=XY\cap PQ$. Due to the angle condition $\angle MPQ=\angle AQM$, we get that $PQXY$ is cyclic. Let $S'$ be the center of $(PQXY)$. Let the tangents to $(PQXY)$ at $P,Q$ meet at $R$. Let $N=AM\cap (MQX)$. [asy][asy] size(15cm); import olympiad; void my_dot(pair A) { fill(circle(A, .02),black); } pair A=dir(130); pair B=dir(206); pair C=dir(334); pair P=(3B+1A)/(3+1); pair M=(1B+1C)/(1+1); pair X=extension(P,M,A,C); pair G=foot(P,A,M); pair H=(2G-1P)/(2-1); pair I=(2M-1P)/(2-1); pair J=foot(I,B,C); pair K=(2J-1I)/(2-1); pair Q=intersectionpoints(circumcircle(A,H,M),circumcircle(M,K,C))[1]; pair Y=extension(Q,M,A,B); pair N=intersectionpoints(circumcircle(P,M,Y),circumcircle(Q,M,X))[0]; pair Z=extension(P,Q,X,Y); pair S=extension(M,rotate(90,M)*C,Z,N); pair L=extension(A,M,P,Q); pair U=(1P+1Q)/(1+1); pair R=extension(S,U,A,M); draw(A--B--C--A,red); draw(B--Y,red); draw(C--X,red); draw(A--M--N,red); draw(circumcircle(P,Q,N),blue); draw(Z--N--S,orange); draw(circumcircle(P,Q,X),purple); draw(arc(circumcenter(X,N,Y),circumradius(X,N,Y),100,150),dashed+green); draw(P--Q--Z,orange); draw(A--Z,orange); draw(Y--X--Z,orange); draw(P--M--X,red); draw(Q--M--Y,red); draw(S--P,orange); draw(S--Q,orange); draw(N--P,orange); draw(N--Q,orange); draw(R--P,orange); draw(R--Q,orange); label("$A$",A,dir(160)); label("$B$",B,dir(300)); label("$C$",C,dir(30)); label("$P$",P,dir(150)); label("$M$",M,dir(260)); label("$X$",X,dir(310)); label("$Q$",Q,dir(70)); label("$Y$",Y,dir(340)); label("$R$",R,dir(60)); label("$S$",S,dir(290)); label("$N$",N,dir(330)); label("$Z$",Z,dir(0)); my_dot(A); my_dot(B); my_dot(C); my_dot(P); my_dot(M); my_dot(X); my_dot(Q); my_dot(Y); my_dot(N); my_dot(Z); my_dot(S); my_dot(L); my_dot(R); [/asy][/asy] Claim 1. $R$ is the midpoint of arc $PQ$ in $(PQN)$. Proof. By Pascal on $PPXQQY$ we get that $A-R-M$ are collinear. Considering the inversion of radius $\sqrt{AP\cdot AY}$, we get that $PYMN$ is also cyclic. Now, $$ \measuredangle QPR=\measuredangle QYP=\measuredangle QXP=\measuredangle QXM=\measuredangle QNM=\measuredangle QNR \Longrightarrow R\in (PNQ) $$Also $RP=RQ$, which proves the Claim. $\square$ Claim 2. $S'\in (PRQN)$. Proof. Since $S'$ is the center of $(PQXY)$, then $S'$ lies on the perpendicular bisector line of $\overline{PQ}$. Also, $$ \measuredangle QS'P=2\cdot \measuredangle QXP=\measuredangle QXP+\measuredangle QYP=\measuredangle QXM+\measuredangle MYP=\measuredangle QNM+\measuredangle MNP=\measuredangle QNP $$The Claim is true, and indeed $(PRQNS')=(\overline{RS'})$. $\square$ Claim 3. $S'-N-Z$ are collinear. Proof. Note that $S'NXY$ is cyclic, because $$ \measuredangle S'NX=\measuredangle RNX-\measuredangle RNS=\measuredangle MNX-90^o=(180^o-\measuredangle XPM)-90^o=90^o-\measuredangle XPY=\frac{180^o-\measuredangle XS'Y}{2}=\measuredangle S'YX $$By radical axes on $(PQXY)$,$(PQS'N)$,$(S'NXY)$ we reach the Claim. $\square$ Claim 4. $S\equiv S'$. Proof. Since $PRQS'$ is a kite, then $$ -1=(P,Q;R,S')\stackrel{(N)}=(P,Q;NR\cap PQ,Z)\stackrel{(A)}=(B,C;M,AZ\cap BC) \Longrightarrow AZ\parallel BC $$Now, by Brocard on $PQXY$ we get that the polar of $M$ wrt $(PQXY)$ is line $AZ$. Therefore, $$ S'M\perp AZ $$Hence $S'M\perp BC$, meaning that $S'$ lies on the perpendicular bisector line of $\overline{BC}$. $\square$ By Claim 2 and Claim 4, we get $$ \measuredangle QSP=\measuredangle QNP=360^o-\measuredangle NPA-\measuredangle AQN-\measuredangle PAQ=((180^o-\measuredangle NPA)+(180^o-\measuredangle AQN))-\measuredangle BAC= $$$$ =(\measuredangle YPN+\measuredangle NQX)-\measuredangle BAC=(\measuredangle YMN+\measuredangle NMX)-\measuredangle BAC=\measuredangle YMX-\measuredangle BAC=\measuredangle QMP-\measuredangle BAC \thickspace \blacksquare $$