Since the cross ratio $(B,C,D,P)$ is harmonic, then $MB^2=MC^2=MD \cdot MP$ $\Longrightarrow$ circle $(M)$ with diameter $\overline{BC}$ is orthogonal to $(K) \equiv \omega$ $\Longrightarrow$ $Q \in (M).$ Inversion with center $B$ and power $BD \cdot BP$ takes $(K)$ into itself and carries $(M)$ into a line orthogonal to $(K),$ due to conformity. $R$ is the inverse of $Q,$ hence $RK$ is perpendicular bisector of $\overline{DP}$ $\Longrightarrow$ $\triangle SDP$ is S-isosceles $\Longrightarrow$ $\angle SDP=\angle SPD \equiv \angle QPD=\angle MQD$ $\Longrightarrow$ $SD$ is tangent to $\odot(QDM).$

Other solution, thanks to the same brilliant idea of Luis: $Q$ belongs to Apollonius circle of $\Delta ABC$, hence $QD, QP$ are bisectors of $\angle BQC$ which, being onto the circle of diameter $BC$, is a right angle, consequently $\angle BQD=45^\circ$ and $R$ is the midpoint of the arc $DP$ of the circle $(ADP)$ and $\angle SDP=\angle SPD=\angle DQM$, done. Best regards, sunken rock

Pretty standard problem.

Wow, G13! To be honest I found the G4 more challenging than this. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -20.45084917598212, xmax = 16.15459252857807, ymin = -9.314976474090676, ymax = 12.37362934827489; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((0.9200000000000010,5.640000000000006)--(-0.6600000000000008,2.340000000000002)--(4.280000000000005,1.440000000000002)--cycle, zzttqq); /* draw figures */ draw((0.9200000000000010,5.640000000000006)--(-0.6600000000000008,2.340000000000002), zzttqq); draw((-0.6600000000000008,2.340000000000002)--(4.280000000000005,1.440000000000002), zzttqq); draw((4.280000000000005,1.440000000000002)--(0.9200000000000010,5.640000000000006), zzttqq); draw((0.9200000000000010,5.640000000000006)--(-11.16897470504708,4.254590533308175)); draw((-11.16897470504708,4.254590533308175)--(-0.6600000000000008,2.340000000000002)); draw((0.9200000000000010,5.640000000000006)--(1.339939757288566,1.975638505757146)); draw(circle((-4.914517473879258,3.115114519532661), 6.357408343222671)); draw((1.810000000000002,1.890000000000002)--(0.4841939791729930,-0.2420502796929700)); draw((-3.775041460103734,9.369571750700484)--(0.4841939791729930,-0.2420502796929700)); draw((-11.16897470504708,4.254590533308175)--(0.4841939791729930,-0.2420502796929700)); draw((-3.775041460103734,9.369571750700484)--(-5.131368636470008,1.924842582645597)); draw((-5.131368636470008,1.924842582645597)--(1.339939757288566,1.975638505757146)); draw(circle((1.349969878644284,0.6978192528785712), 1.277858617516652)); /* dots and labels */ dot((0.9200000000000010,5.640000000000006),dotstyle); label("$A$", (1.095691542743477,5.933344649930528), NE * labelscalefactor); dot((-0.6600000000000008,2.340000000000002),dotstyle); label("$B$", (-0.4670245972959622,2.618492231665050), NE * labelscalefactor); dot((4.280000000000005,1.440000000000002),dotstyle); label("$C$", (4.457898995555603,1.718746575278705), NE * labelscalefactor); dot((-11.16897470504708,4.254590533308175),dotstyle); label("$P$", (-10.97984226665219,4.560048648077688), NE * labelscalefactor); dot((1.339939757288566,1.975638505757146),dotstyle); label("$D$", (1.521886853663324,2.239651955291852), NW * labelscalefactor); dot((1.810000000000002,1.890000000000002),dotstyle); label("$M$", (1.995437199129820,2.192296920745202), NE * labelscalefactor); dot((0.4841939791729930,-0.2420502796929700),dotstyle); label("$Q$", (0.6694962318236298,0.06132036614596580), E * labelscalefactor); dot((0.4841939791729930,-0.2420502796929700),dotstyle); dot((-3.775041460103734,9.369571750700484),dotstyle); label("$R$", (-3.592456877374840,9.674392379115854), NE * labelscalefactor); dot((-5.131368636470008,1.924842582645597),dotstyle); label("$S$", (-4.918397844681031,2.192296920745202), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\omega$ denote the circumcircle of $\triangle APD.$ By the angle bisector theorem, $\dfrac{BD}{DC} = \dfrac{AB}{AC} = \dfrac{BP}{PC},$ so $\omega$ is the Appolonius circle of $B$ and $C$ with ratio $\dfrac{AB}{AC}.$ Since $Q$ lies on this circle too, $\dfrac{QB}{QC} = \dfrac{BD}{DC},$ implying $QD$ bisects $\angle BQC$ internally. Since $\angle PAD = 90^{\circ}$ and quadrilateral $APQD$ is cyclic, $\angle PQD = 90^{\circ},$ so $QP$ bisects $\angle BQC$ externally (this follows from the fact that the internal and external angle bisectors are perpendicular). Also, note that since $DA \perp AP,$ $AD$ is a diameter of $\omega.$ Now, note that $PB \times DC = PC \times BD,$ which implies $(P,D,B,C)$ is harmonic. Hence, $MC^2 = MB^2 = MD \cdot MP = MQ^2,$ where we have used the fact that $MQ^2 = MD \times MP$ which follows from PoP. Hence, $MB=MC=MQ,$ which implies that $M$ is the circumcenter of $\triangle BCQ.$ Since $M$ is the midpoint of $BC,$ this forces $\triangle BQC$ to be right angled at $Q.$ By our previous observations, we have that $\angle BQM = \angle QMC = \angle BQP = 45^{\circ}.$ Now, $\angle RPD = \angle PQR = 45^{\circ} = \angle RQD = \angle RDP,$ so $\triangle RPD$ is isoceles with $RP=RD.$ Since $AD$ is a diameter of $\omega$ and $R$ lies on the perpendicular bisector of $PD,$ $RS$ must bisect $PD,$ implying $RS$ is the perpendicular bisector of $PD.$ Hence, $SP = SD$ and $\angle SDP = \angle SPD.$ Since $MQ$ is tangent to $\omega,$ $\angle MQD= \angle QPD.$ It follows that $\angle SDP = \angle MQD,$ completing the proof. $\blacksquare$

Obviously, the circle with centre $M$ through $B, C$ is orthogonal to the circle $APD$, so $\angle BQC = \dfrac{\pi}{2}$, thus $QB$ bisects $\angle PQD$. So, $SDP$ is isosceles, or $\angle SDP = \angle SPD = \angle MQD$...

Yes its just an easy cross ratio exercise. $AP,AD,AB,AC$ being a harmonic pencil implies $(P,D;B,C)=-1 \implies MD \cdot MP=MB^2=MQ^2=MC^2 \implies QB \perp QC$.Also $Q$ being a point on the Apollonius circle means $\angle{BQD}=\angle{{DQC}=45^{\circ} \implies PR=RM \implies \angle{DQM}=\angle{QPD}=\angle{SMP}}$ which implies the result.

TBH This was really very easy. ELMO 2013 Shortlist G13 wrote: In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. Proposed by Ray Li Solution:- Claim 1:- $(P,D;B,C)=-1$. Just see that $\angle BAD=\angle DAC$ and $\angle PAD=90^\circ$. So, $(P,D;B,C)=-1$. Claim 2:- $MB=MQ=MC$. As $(P,D;B,C)=-1$ and $MB=MC\implies MD.MP=MB^2=MC^2=MQ^2\implies \angle BQC=90^\circ$. Claim 3:- $R$ is the midpoint of $\widehat{PD}$ of $\odot(APD)$. As $(P,D;B,C)=-1\implies \angle PQB=\angle BQD$ as $\angle BQC=90^\circ$. So, $R$ is the midpoint of $\widehat{PD}$ of $\odot(PQD)$. Claim 4:- $\angle SDQ=\angle QMD$. Let $\angle QPC=\theta\implies \angle QSD=2\theta\implies\angle QDS=90^\circ-2\theta$. Now notice that $\angle PQB=45^\circ\implies\angle QBM=45^\circ-\theta\implies\angle QMD=90^\circ-2\theta=\angle QDS$. So, $SD$ is tangent to $\odot(QDM)$. $\blacksquare$.

v_Enhance wrote: In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. Proposed by Ray Li Wut? (did in around 5 mins coz i'm stupid, how could i forgot cross ratio+angle bisector stuff) No need of even naming claims n'stuff, just note that $-1=(P, D; B, C)$ so since $\angle PQD=90$ we have that $\angle BQD=\angle DQC$ and since using MacLaurins theorem and PoP we have $MQ^2=MD \cdot MP=MB^2=MC^2$ so $M$ is center of $(BQC)$ meaning that $\angle BQC=90$ to $\angle PQB=\angle BQD=\angle DQC=45$ meaning that $PS=SD$ as $S$ lies on the perpbisector of $PD$, so now by angle chase $$\angle DQM=\angle SPD=\angle SDP \implies SD \; \text{tangent to} \; (QDM)$$Thus we are done

... It is well-known that $MD\cdot MP=MB^2=MC^2$, and by power of a point it also equals $MQ^2$, hence $M$ is the center of $(BCQ)$, i.e. $\angle BQC=90^\circ$. Since $Q$ lies on the $A$-apollonius circle $\omega$ of $\triangle ABC$, it also follows that $\overline{QD}$ bisects $\angle BQC$ internally, hence $\angle BQD=\angle RQD=45^\circ$. Since $\overline{PD}$ is a diameter of $\omega$, it follows that $\overline{RS}$ bisects $\overline{DP}$. Now, we have $\angle SDP=\angle SPD=90^\circ-\angle BDQ$ and $\angle DQM=\angle BQM-\angle BQD=\angle BQM-45^\circ$, so it suffices to show that $135^\circ=\angle BQM+\angle BDQ=\angle QBD+\angle BDQ$, which is clear by looking at $\triangle BDQ$. $\blacksquare$

Note that $MQ^2 = MD\cdot MP = MB^2=MC^2$ so $Q$ lies on $(BC)$ and hence $R' = CQ\cap (PD)$ is diametrically opposite to $R$. But by projecting through $Q$ onto $BC$, $(PD;RR')$ is a harmonic bundle, so $RR'$ is actually the diameter perpendicular to $PD$. Then we have $$\measuredangle SDM = \measuredangle DPS = \measuredangle DPQ = \measuredangle DQM$$ as desired.