$AB\cap QS=T$ We will show that $T,D,X$ are collinear $\iff \frac{TS}{TK}=\frac{SD}{KX}$ Let $\angle CRQ=\alpha$ and $\angle QPB=\beta$. We have $QX.QR=QK.QS=QP^2$ \[\frac{TS}{TK}=\frac{AS}{PK}=\frac{cos\beta .\frac{PS}{sin 2\beta}}{cos\beta.PQ}=\frac{PS}{PQ}.\frac{1}{sin 2\beta}\]\[\frac{SD}{KX}=\frac{cos\alpha.\frac{SR}{sin 2\alpha}}{cos \alpha.XQ}=\frac{SR}{XQ}.\frac{1}{sin 2\alpha}=\frac{SR}{\frac{QP^2}{QR}}.\frac{1}{sin 2\alpha}=\frac{SR.RQ}{QP^2}.\frac{1}{sin 2\alpha}\]$\frac{PS}{PQ}.\frac{1}{sin 2\beta}\overset{?}{=}\frac{SR.RQ}{QP^2}.\frac{1}{sin 2\alpha} \iff \frac{sin 2\alpha}{sin 2\beta}\overset{?}{=}\frac{SR}{SP}.\frac{QR}{QP}$ \[\frac{SR}{SP}.\frac{QR}{QP}=\frac{cos \alpha}{cos \beta}.\frac{sin \alpha}{sin \beta}=\frac{2cos\alpha.sin\alpha}{2.cos\beta.sin \beta}=\frac{sin 2\alpha}{sin 2\beta}\]As desired.$\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(16.12338664364587cm); 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 = -28.323118931360163, xmax = 27.800267712285713, ymin = -12.066771626573988, ymax = 16.89801890572312; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw(circle((0.,0.), 6.), linewidth(0.4) + red); draw((-5.909431225761404,1.0385675654458277)--(-0.023632847618183822,5.999953457195602), linewidth(0.4) + qqwuqq); draw((-0.03900421829849259,9.902467032588484)--(-7.151400313586333,-6.028214753981295), linewidth(0.4) + qqwuqq); draw((2.9486386462174736,6.0116607545596255)--(0.023632847618183822,-5.999953457195602), linewidth(0.4)); draw((-5.909431225761404,1.0385675654458277)--(-14.326433474790848,-6.056476050766989), linewidth(0.4)); draw((-3.5484059866102102,14.472759752125091)--(-7.151400313586333,-6.028214753981295), linewidth(0.4)); draw((-3.5484059866102102,14.472759752125091)--(12.135393688838272,-5.9522471874069325), linewidth(0.4)); draw((4.758866145569481,3.654202102860852)--(-8.071659031999236,3.6036647364386), linewidth(0.4) + qqwuqq); draw((2.9486386462174736,6.0116607545596255)--(-5.040953420178436,5.980191041805413), linewidth(0.4)); draw((-14.326433474790848,-6.056476050766989)--(12.135393688838272,-5.9522471874069325), linewidth(0.4)); draw((-8.071659031999236,3.6036647364386)--(-0.03900421829849259,9.902467032588484), linewidth(0.4) + linetype("2 2") + ffxfqq); draw((-0.03900421829849259,9.902467032588484)--(0.023632847618183822,-5.999953457195602), linewidth(0.4)); draw((-5.909431225761404,1.0385675654458277)--(0.023632847618183822,-5.999953457195602), linewidth(0.4)); /* dots and labels */ dot((0.,0.),linewidth(4.pt) + dotstyle); label("$I$", (-0.5405647473201336,-0.5728071296306912), NE * labelscalefactor); dot((-0.023632847618183822,5.999953457195602),dotstyle); label("$S$", (0.3132726153099383,6.356411467097982), NE * labelscalefactor); dot((0.023632847618183822,-5.999953457195602),linewidth(4.pt) + dotstyle); label("$Q$", (-0.770444037258999,-6.812387856542767), NE * labelscalefactor); dot((4.758866145569481,3.654202102860852),dotstyle); label("$P$", (4.878018515524553,3.991938770583932), NE * labelscalefactor); dot((-5.909431225761404,1.0385675654458277),dotstyle); label("$R$", (-5.499389430287089,0.7407888128771141), NE * labelscalefactor); dot((2.9486386462174736,6.0116607545596255),linewidth(4.pt) + dotstyle); label("$A$", (3.071824094576324,6.290731669972591), NE * labelscalefactor); dot((12.135393688838272,-5.9522471874069325),linewidth(4.pt) + dotstyle); label("$B$", (12.266995692130944,-5.695831305411132), NE * labelscalefactor); dot((-7.151400313586333,-6.028214753981295),linewidth(4.pt) + dotstyle); label("$C$", (-7.765342431113049,-5.794351001099217), NE * labelscalefactor); dot((-5.040953420178436,5.980191041805413),linewidth(4.pt) + dotstyle); label("$D$", (-5.729268720225955,6.323571568535287), NE * labelscalefactor); dot((-8.071659031999236,3.6036647364386),linewidth(4.pt) + dotstyle); label("$X$", (-8.192261112428085,4.221818060522798), NE * labelscalefactor); dot((-0.03900421829849259,9.902467032588484),linewidth(4.pt) + dotstyle); label("$Y$", (0.5103120066861087,9.67324122193019), NE * labelscalefactor); dot((-3.5484059866102102,14.472759752125091),linewidth(4.pt) + dotstyle); label("$E$", (-4.3828328791554565,13.942428035080557), NE * labelscalefactor); dot((-2.841964553747014,3.624263658342897),linewidth(4.pt) + dotstyle); label("$F$", (-2.7408379510207035,2.973901915140383), NE * labelscalefactor); dot((-1.784412870601648,5.993018029067838),linewidth(4.pt) + dotstyle); label("$J$", (-2.2482394725802775,6.5206109599114574), NE * labelscalefactor); dot((-14.326433474790848,-6.056476050766989),linewidth(4.pt) + dotstyle); label("$T$", (-14.464681737902843,-5.630151508285742), NE * labelscalefactor); dot((-0.01431925855480821,3.635401296468975),linewidth(4.pt) + dotstyle); label("$L$", (0.11623322393376784,3.8934190748958466), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We solve for case $AD<BC$ other case work similarly. Let $I$ we center of $\omega$.All pole and polar are taken with respect to $\omega$ and we use well know La Hire’s Theorem. Let $Y=QS \cap AB$.Note $\overline{Q-I-S}$ and $QS \perp PX$ hence from $YP$ is tangent to $\omega$ and $IY \perp PX$ we get $PX$ is polar of $Y$ with respect to $\omega$. As $X=QR \cap PX$ polar of $X$ is pass through pole of $QR$ and $PX$ which are $C,Y$ respectively.Hence $CY$ is polar of $X$ We already know $RS$ is polar of $D$. Proving $X,D,Y$ are collinear is same proving as $RS,CY,PX$ are concurrent. Define $F = CY \cap RS$,$J=AD \cap CY$ and $T=RS \cap BC$. We claim $(Y,F;J,C)=-1$ Note from $SQ$ is diameter so we get $\measuredangle{QRS}=\measuredangle{TRQ}=90$ and $CR=CQ$.hence $C$ is center of $\triangle TRQ$ which mean $C$ is midpoint of $TQ$. $$(T,Q;C,\infty{\overline{BC}})\stackrel{S}{=}(F,Y;C,J)=-1$$ From observing $PX$ is polar of $Y$ we get $(Y,L;S,Q)=-1$ where $L=PX \cap IY$ if $F' = PX \cap CY$.From $JS \parallel CQ \parallel PF'$ we get $\triangle YSJ \sim \triangle YLF' \sim \triangle YQC$ which mean length ratio are also same hence $\Rightarrow$ $$(Y,F;J,C)=(Y,F';J,C)=-1$$but from claim we get $F'=F$ hence $AF\parallel BC$ which give $RS,CY,PX$ are concurrent.hence $X,D,Y$ are collinear.

Obviously $QS$ is diameter. Let $T$ be reflection of P over $QS$. Then just do Pascal to $SPPRTT$, $PPTSQR$, $SSTRRP$.

great good solve

Nice! Let $E = QS \cap PX$, and $\infty$ be the point at infinity along line $BC$. We want to show that $AS:SD = PE:EX$, which is well-known to be equivalent to $CQ : QB = PE : EX$. But \[ (P, E; X, \infty) \stackrel{Q} = (P,S;R,Q)_\omega = (AB, AD; DC, BC)_\omega = (B, \infty; C, Q) = (C, Q; B, \infty) \]where we have used the concept of the cross ratio of four tangents to a conic.