Easy problem.Similar ideas are used also in 2006 G2 and 2007 G3. Let's apply an homothety with center P which takes $A$ to $D$ and $B $ to $C $.Let it take $Q $ to $Q'$.Let $\angle QDC=\alpha $ , $\angle QDA=\beta $ ,$\angle QCD=x $ and $\angle QQ'D=2z $ .We have to prove that $z=x $. Let the angle bisector of $QQ'D $ meet $CD $ at $I $.It is easy to see that $I $ is incenter of $QQ'D $.Clear that $\angle IQC=90+z-x $ and $\angle IQ'C=90-z+x $ so $IQCQ'$ is cyclic. Then $\angle QQ'I=z=\angle QCI=x $ as desired.

Let $O$ be the circumcenter of $\triangle QCD$ and $\omega$ be the circumcircle of $\triangle ODQ$. $CO$ cut $\omega$ again at $S$. Since $\angle OCQ=\angle OQC=\angle BQC=90^\circ -\angle QDC\Longrightarrow QB\parallel CO$ , since $\angle QOS=180^\circ -2\angle QDC\Longrightarrow \angle CDS=\angle QAB\Longrightarrow \triangle DCS$ and $\triangle ABQ$ are homothetic $\Longrightarrow A,Q,S$ lie on a line $\Longrightarrow \angle PQA=\angle PSD=\angle QOD=2\angle QCD$. [asy][asy] import graph; size(13.60992169446671cm); 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 = -14.843781433163949, xmax = 38.766140261302766, ymin = -9.131333541801133, ymax = 16.030787903061157; /* image dimensions */ draw(arc((1.2546591732673007,5.624816236995388),0.8424817448056582,-45.50311327983149,1.7887540200298102)--(1.2546591732673007,5.624816236995388)--cycle); draw(arc((-0.672859921619772,-0.3957708050867718),0.8424817448056582,1.7887540200298055,49.08062131989111)--(-0.672859921619772,-0.3957708050867718)--cycle); draw(arc((-0.672859921619772,-0.3957708050867718),1.54454986547704,-45.50311327983149,1.788754020029804)--(-0.672859921619772,-0.3957708050867718)--cycle); draw(arc((3.003294456888648,3.84519862753241),0.8424817448056582,-23.14189936985122,19.566233330287503)--(3.003294456888648,3.84519862753241)--cycle); draw(arc((3.003294456888648,3.84519862753241),1.4041362413427636,-65.85003206998992,-23.141899369851217)--(3.003294456888648,3.84519862753241)--cycle); draw(arc((12.,0.),0.8424817448056582,156.85810063014878,199.56623333028747)--(12.,0.)--cycle); /* draw figures */ draw((3.003294456888648,3.84519862753241)--(1.2546591732673007,5.624816236995388)); draw((3.003294456888648,3.84519862753241)--(-0.672859921619772,-0.3957708050867718)); draw((3.003294456888648,3.84519862753241)--(8.661166826164914,5.856119945455693)); draw((3.003294456888648,3.84519862753241)--(12.,0.)); draw((-0.672859921619772,-0.3957708050867718)--(1.2546591732673007,5.624816236995388)); draw((3.9654890226445043,14.092066760086112)--(1.2546591732673007,5.624816236995388)); draw((3.9654890226445043,14.092066760086112)--(8.661166826164914,5.856119945455693)); draw((8.661166826164914,5.856119945455693)--(12.,0.)); draw((12.,0.)--(-0.672859921619772,-0.3957708050867718)); draw((1.2546591732673007,5.624816236995388)--(8.661166826164914,5.856119945455693)); draw((3.9654890226445043,14.092066760086112)--(3.003294456888648,3.84519862753241)); draw(circle((5.727018303415523,-2.229543509872047), 6.657414184809026)); draw((3.003294456888648,3.84519862753241)--(2.319131856673411,-3.440774703779689)); draw((2.319131856673411,-3.440774703779689)--(-0.672859921619772,-0.3957708050867718)); draw(circle((2.953280044813141,0.17478649207755234), 3.670752877205104)); draw((5.727018303415523,-2.229543509872047)--(3.003294456888648,3.84519862753241)); draw((5.727018303415523,-2.229543509872047)--(12.,0.)); draw(arc((3.003294456888648,3.84519862753241),0.8424817448056582,-23.14189936985122,19.566233330287503)); draw(arc((3.003294456888648,3.84519862753241),0.7020681206713818,-23.14189936985122,19.566233330287503)); draw(arc((3.003294456888648,3.84519862753241),1.4041362413427636,-65.85003206998992,-23.141899369851217)); draw(arc((3.003294456888648,3.84519862753241),1.2637226172084872,-65.85003206998992,-23.141899369851217)); draw(arc((12.,0.),0.8424817448056582,156.85810063014878,199.56623333028747)); draw(arc((12.,0.),0.7020681206713818,156.85810063014878,199.56623333028747)); draw((5.727018303415523,-2.229543509872047)--(2.319131856673411,-3.440774703779689)); /* dots and labels */ dot((8.661166826164914,5.856119945455693),linewidth(2.pt) + dotstyle); label("$B$", (8.97036922000932,5.977172415046983), NE * labelscalefactor); dot((3.003294456888648,3.84519862753241),linewidth(2.pt) + dotstyle); label("$Q$", (1.7531089395075168,4.123712576474538), NE * labelscalefactor); dot((12.,0.),linewidth(2.pt) + dotstyle); label("$C$", (12.42454437371252,-0.2291097716880237), NE * labelscalefactor); dot((1.2546591732673007,5.624816236995388),linewidth(2.pt) + dotstyle); label("$A$", (0.320889973337898,5.8367587909127066), NE * labelscalefactor); dot((-0.672859921619772,-0.3957708050867718),linewidth(2.pt) + dotstyle); label("$D$", (-1.6729834893688262,-0.9030951675325494), NE * labelscalefactor); dot((3.9654890226445043,14.092066760086112),linewidth(2.pt) + dotstyle); label("$P$", (3.7188996773873857,14.76706528585267), NE * labelscalefactor); dot((5.727018303415523,-2.229543509872047),linewidth(2.pt) + dotstyle); label("$O$", (6.134014012496939,-3.037382254373547), NE * labelscalefactor); dot((2.319131856673411,-3.440774703779689),linewidth(2.pt) + dotstyle); label("$S$", (1.8935225636417932,-4.301104871582033), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

let <QAB=<QDC=x and <QBA=y and <QCD=z. we have <BQC=y+z. so <BQC +<QAB= x+y+z =90. drew parallel to DQ from A cut PQ to P'. so PP'/P'Q=PA/AD. AB is parallel to DC . so PA/AD=PB/BC. so PP'/P'Q=PB/BC. so P'B is parallel to QC . so <P'AB=x and <P'BA=z.(we want to prove z=<PQA/2) so AB is bisector of <P'AQ. bisector's of 180-<AP'Q and 180-AQP' meet at A' in AB. so <P'A'A=<P'QA/2 and <AA'Q =<AP'Q/2 . so <P'A'A+<QA'A+<P'AQ=<P'A'Q+<P'AQ=90+x and <P'BQ+<P'AQ=90+x. so <P'A'Q=<P'BQ_<BP'A'_<BQA'=<P'BQ. so <BP'A'=<BQA'=0. so A' is B . so we have <P'A'A=<P'QA/2. so we have <P'BA=<P'QA/2=<PQA/2=z

Let $Q'$ be the reflection of $Q$ about $CD$ . so we have $DQ'||QA$ . It's easy to see that $\angle{CQQ'}+\angle{AQB}=180 $ , so $\frac{\sin{AQB}}{\sin{DQC}}=\frac{\sin{CQQ'}}{\sin{CQD}} $ $\frac{\sin{AQB}}{\sin{DQC}}=\frac{AB}{CD} . \frac{QC}{QD} = \frac{PB}{PC}.\frac{QC}{QB} = \frac{\sin{PQB}}{\sin{PQC}}$ $\rightarrow $$\frac{\sin{CQQ'}}{\sin{CQD}}= \frac{\sin{PQB}}{\sin{PQC}}$ and we know $\angle{PQB} - \angle{PQC} = \angle{CQQ'} - \angle{CQD}$ so we have $\angle{PQC} =\angle{CQD}$ so if $L$ is intersection of $DQ' , PQ$ we have $QLQ'C$ cyclic so $2\angle{QCD} = \angle{QCQ'} = \angle{QLD} = \angle{PQA}$ and the problem is finished $\blacksquare$

Awesome problem; thanks to Aryan-23 for the rec. [asy][asy] size(200); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,5), B = (6,5), C = (9,0), D = (-1,0), P = extension(A,D,B,C); pair Q = (1,2.8); pair Xp = rotate(90,D)*C; path circ = circumcircle(Q,D,C); pair O = circumcenter(Q,D,C); pair X = extension(D,Xp,Q,C), Y =extension(D,Xp,O,C), Qp = extension(C,Y,P,Q); draw(A--B--C--D--A,red); draw(A--Q--B^^C--Q--D^^C--Qp--D,blue); draw(Q--X--Y--Qp,purple); draw(circumcircle(Q,D,C),rgb(0.1,0.4,0.1)); draw(circumcircle(Q,D,Qp),gray(0.5)+linetype("3 3")); dot("$A$",A,NW); dot("$B$",B,NE); dot("$C$",C,E); dot("$D$",D,W); dot("$X$",X,NW); dot("$Y$",Y,SW); dot("$Q$",Q,dir(80)); dot("$Q'$",Qp,S); dot("$O$",O,SE); clip((-3,10)--(-3,-4)--(10,-4)--(10,10)--cycle); [/asy][/asy] Construct point $Q'$ outside $ABCD$ such that $\triangle ABQ\sim\triangle DCQ'$. Then $\angle Q'DC = \angle QDC =: \alpha$ and $\angle QCQ' = \angle BQQ' = 90^\circ - \alpha$. Let the perpendicular to $CD$ through $D$ intersect $CQ$ at $X$ and $CQ'$ at $Y$. Note that \[ \angle XDQ = 90^\circ - \alpha = \angle QCQ', \]so quadrilateral $QDYC$ is cyclic; this means the circumcenter $O$ of $\triangle QDC$ lies on $\overline{CY}$. In turn, \[ \angle QOQ' = 2\angle QCQ' = 2(90^\circ - \alpha) = 180^\circ - \angle QDQ', \]so $QDQ'O$ is cyclic. As a result, \[ \angle PQA = \angle PQ'D = \angle QOD = 2\angle QCD. \quad\blacksquare \]

Let $Q'$ be such that $Q,Q'$ are on different sides of $CD$ and $QAB$ and $Q'DC$ are similar then by Homothety we have $P,Q,Q'$ are collinear so $\angle PQA = \angle QQ'D$ so we need to prove $\angle QQ'D = 2\angle QCD$. Let angle bisector of $QQ'D$ meet $CD$ at $S$. Note that $S$ is incenter of $QDQ'$. $\angle QSQ' = \angle 90 + \angle QDC $ and $\angle QCQ' = \angle BQC = \angle 90 - \angle QCD$ so $QSQ'C$ is cyclic so $\angle QCD = \angle QCS = \angle QQ'S = \frac{QQ'D}{2} = \frac{PQA}{2}$. we're Done.

Let $Q'$ be the reflection of $Q$ over $CD$. $\angle{Q'DC}=\angle{CDQ}=\angle{QAB}$ and therefore, $Q'D$ is parallel to $QA$. Let $T = PQ \cap DQ'$. Notice that $\triangle{DTC}$ and $\triangle{AQB}$ are homothetic, thus $TC$ is parallel to $QB$, which implies that $\angle{QCT} = \angle{CQB} = 90^{\circ} - \angle{Q'DC} = \angle{DQ'Q}$, therefore, $QCTQ'$ is cyclic $\implies \angle{PQA} = \angle{QTQ'} = \angle{QCQ'} = 2\angle{QCD}$.