Very interesting problem! The concyclicity of points $A,P,Q,R $ is not necessary and I do not use it below: Let $ R \equiv QS \cap PD $. It is equivalent to prove that $ A,B,R $ are collinear. First of all, it is obvious that $ AD $ is a symmedian and that $ AT $ is tangent to the circumcircle $ (\omega) $ of $ \triangle ABC $. Let $ J $ be the intersection point of the tangents to $(\omega) $ at $ B,C $ and $ F \equiv AJ \cap (\omega) $. From La Hire's Theorem, $ AF $ is the polar of $ T $ so $ (C,B;T,D) = -1 $ Thus, $ \frac{BD}{DC} = \frac{TB}{TC} = \frac{c^2}{b^2} $. Observe that $ \frac{SB}{SD} = \frac{\frac{TB+BD}{2} - BD}{\frac{TD}{2}} = \frac{TB-BD}{TD} = \frac{\frac{c^2}{b^2}TC - \frac{c^2}{b^2}DC}{TD} = \frac{c^2}{b^2} $. It follows that $ \frac{SB}{SD} = \frac{BD}{DC} \implies \frac{SB}{BD} = \frac{SD}{DC} ~(1)$. Applying Menelaus Theorem at $ \triangle QCS $ we get $ \frac{CP}{PQ}\cdot \frac{SD}{DC} \cdot \frac{QR}{SR} = 1~(2) $. Furthermore, $ AP \parallel DC \implies \frac{CP}{PQ} = \frac{DA}{AQ}~(3) $. Consequently, $ (1),(2),(3) \implies \frac{DA}{AQ} \cdot \frac{SB}{BD} \cdot \frac{QR}{SR} = 1 $. From reverse Menelaus Theorem at $ \triangle SDQ $ it follows that $ A,B,R $ are collinear and this completes the proof. With Best Regards, Grigoris

As Grigoris mentioned, $A, P, Q, R$ need not be concyclic for $Q, R, S$ to be collinear. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(400); 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 = -7.000000000000000, xmax = 23.00000000000000, ymin = -15.50000000000002, ymax = 4.000000000000000; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((4.980000000000006,2.440000000000003)--(14.78000000000002,-5.540000000000006)--(2.900000000000003,-5.500000000000006)--cycle, zzttqq); /* draw figures */ draw((4.980000000000006,2.440000000000003)--(14.78000000000002,-5.540000000000006), zzttqq); draw((14.78000000000002,-5.540000000000006)--(2.900000000000003,-5.500000000000006), zzttqq); draw((2.900000000000003,-5.500000000000006)--(4.980000000000006,2.440000000000003), zzttqq); draw((4.980000000000006,2.440000000000003)--(8.840000000000010,-5.520000000000006)); draw((4.980000000000006,2.440000000000003)--(0.3289413766219917,-5.491343236958329)); draw((-5.766509295032836,-5.470819834023465)--(2.900000000000003,-5.500000000000006)); draw((4.980000000000006,2.440000000000003)--(21.18778295665536,2.385428340213284)); draw((21.18778295665536,2.385428340213284)--(2.322020852832571,-7.706324244475684)); draw((4.980000000000006,2.440000000000003)--(-5.766509295032836,-5.470819834023465)); draw((21.18778295665536,2.385428340213284)--(7.961389590538268,-13.97355783826812)); draw((4.980000000000006,2.440000000000003)--(7.961389590538268,-13.97355783826812)); draw((0.3289413766219917,-5.491343236958329)--(7.961389590538268,-13.97355783826812)); draw((2.900000000000003,-5.500000000000006)--(2.322020852832571,-7.706324244475684)); /* dots and labels */ dot((4.980000000000006,2.440000000000003),dotstyle); label("$A$", (4.543842585427171,2.938815996269827), NE * labelscalefactor); dot((2.900000000000003,-5.500000000000006),dotstyle); label("$B$", (2.117672547855579,-5.129609942631011), NE * labelscalefactor); dot((14.78000000000002,-5.540000000000006),dotstyle); label("$C$", (15.29234007745946,-6.399117520430095), NE * labelscalefactor); dot((8.840000000000010,-5.520000000000006),dotstyle); label("$M$", (8.747323231917486,-6.455540079443387), NE * labelscalefactor); dot((6.424392048276819,-5.511866639893193),dotstyle); label("$D$", (6.575054709905714,-4.875708427071195), NE * labelscalefactor); dot((-5.766509295032836,-5.470819834023465),dotstyle); label("$T$", (-6.289288745125053,-6.342694961416802), NE * labelscalefactor); dot((0.3289413766219917,-5.491343236958329),dotstyle); label("$S$", (-0.2802862102093663,-6.427328799936741), NE * labelscalefactor); dot((21.18778295665536,2.385428340213284),dotstyle); label("$P$", (21.30134261237514,2.543858083176779), NE * labelscalefactor); dot((7.961389590538268,-13.97355783826812),dotstyle); label("$Q$", (7.985618685238034,-14.80607881341069), NE * labelscalefactor); dot((2.322020852832571,-7.706324244475684),dotstyle); label("$R$", (1.666292075749236,-8.514963483428566), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We use barycentric coordinates. Let $A = (1, 0, 0), B = (0, 1,0), C = (0, 0, 1)$, and let $BC = a, CA = b, AB = c$. Since $\triangle ATB \sim \triangle CTA$, we have $AT^2 = CT \cdot BT$, which implies that $AT$ is tangent to the circumcircle of $\triangle ABC$. It is well-known that the line tangent to the circumcircle at $A$ has equation $c^2y+b^2z = 0$. Intersecting this line with $BC$ gives $T = (0:b^2:-c^2)$. Now $M = \left(0, \tfrac{1}{2}, \tfrac{1}{2}\right)$, and $D$ is the trace of the symmedian point $K = (a^2:b^2:c^2)$ wrt $A$, so $D = (0:b^2:c^2)$. Thus we also have $S = (0:b^4:-c^4)$. Denote by $\mathcal{P}_{\infty}$ the point of infinity of $BC$. Then $\mathcal{P}_{\infty} = (0:-1:1)$. Since $AP\parallel BC$, $\mathcal{P}_{\infty}$ lies on $AP$. Therefore $A\mathcal{P}_{\infty}$ has equation $y+z=0$, and so we can let $P = (1, p, -p)$ for some $p$. Line $PD$ has equation $(-p(b^2+c^2))x+c^2y-b^2z = 0$. $R = AB \cap PD$, so intersecting $PD$ with $z = 0$ yields $R = (c^2:p(b^2+c^2):0)$. In addition, line $PC$ has equation $px-y=0$. Intersecting $PC$ with $AD: c^2y-b^2z = 0$ yields $Q = (b^2:pb^2:pc^2)$. Notice that \begin{align*} \det \left| \begin{array}{ccc}b^2 & pb^2 & pc^2 \\c^2 & p(b^2+c^2) & 0 \\0 & b^4 & -c^4 \end{array} \right| &= b^2(p(b^2+c^2)(-c^4))-pb^2(-c^6)+pc^2(c^2b^4) \\&= -pb^2c^4(b^2+c^2)+pb^2c^4(c^2+b^2) \\&=0\end{align*} implying that $Q, R, S$ are collinear.

We apply the Desargues Involution Theorem to quadrilateral $APRQ$ with respect to its circumcircle (or any conic through the four points) and line $BC$. The the unique involution on line $BC$ which swaps $B,C$ and sends $D$ to itself also swaps the point at infinity along $BC$ and $QR \cap BC$. Since $AT$ is the tangent to $(ABC)$ at $A$, we also have $(T,D;B,C)=-1$. Therefore, the involution we need is the inversion with respect to the circle $(TD),$ since it swaps $B,C$. Since this involution sends the point at infinity to $S$, it follows that $Q,R,S$ collinear.

$A,P,Q,R$ I don't think the cyclical condition is necessary

Let $QR\cap BC=S'$ Claim: $B,C,Q,R$ are cyclic. Proof: $\angle BRQ=180-\angle QPA=180-\angle QCB$ Claim:$S'D$ is tangent to $(DRQ)$. Proof: $\angle RDS'=\angle RPA=\angle RQA=\angle RQD$ \[S'D^2=S'R.S'Q=S'B.S'C\]Also we know that $(T,D;B,C)=-1$ so $S'$ is the midpoint of $DT$ as desired.$\blacksquare$