Since $\measuredangle DYA=\measuredangle BXA=\measuredangle BPC\Longrightarrow APDY$ is cyclic. So from $\measuredangle PYD=\measuredangle PAD=\measuredangle CBP=\measuredangle XAP=\measuredangle PDY$ we conclude that $PY=PD=PC\Longrightarrow \angle YDA=\angle YPA=2\angle YCA$.

I solved using the sine rule and related stuff, pretty easy

The mathematical olympiad holder commitie in Iran said it is wrong in some cases.Here is a picture: http://s6.uplod.ir/i/00920/30a0bapj3ld9.png

We can easily see that $\measuredangle PAX = 180-\measuredangle PBX = \measuredangle PBC$ and since $ABCD$ is isosceles trapezoid, then $ABCD$ is cyclic, we get $\measuredangle PBC = \measuredangle PAD$. So $AP$ is the bisector of $\hat{DAX}$. On the other hand, $\measuredangle CDY = 180-\measuredangle DCB = 180-\measuredangle CDA$. From this, we see that $CD$ is exterior bisector of $\hat{ADY}$. By the equality we got before, we deduce that $C$ is the ex-center of triangle $ADY$ with respect to $A$. So, $\hat {YDA} =2.\hat {YCA} $. Our proof is complete Note: I know it's a bit late but because of the coronavirus, i got nothing else to do

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.080071469145949cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -2.4550944439095383, xmax = 2.0849412906634366, ymin = -1.2787756302270403, ymax = 1.8456668250652017; /* image dimensions */ pen qqqqcc = rgb(0.,0.,0.8); pair A = (-0.5634076280942881,0.8261790632817854), D = (-0.9846332913144251,-0.17463470913687923), B = (0.5555427472464913,0.8314879770518759), C = (0.9862460508917195,-0.16528377748735112), P = (-0.002216324928799256,0.46713088935981495), X = (0.1976529656367213,1.6597482261087875), Y = (-1.1428520606874113,0.19152918840975658); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(1.2)); draw(B--A, linewidth(1.2)); draw(A--D, linewidth(1.2)); draw(D--C, linewidth(1.2)); draw(B--C, linewidth(1.2)); draw(circle((-0.005127315994265423,1.0806753329836478), 0.6135513492533953), linewidth(1.2)); draw(X--B, linewidth(1.2)); draw(A--X, linewidth(1.2)); draw(Y--A, linewidth(1.2)); draw(D--Y, linewidth(1.2)); draw(Y--C, linewidth(1.2)); draw(C--A, linewidth(1.2)); draw(D--B, linewidth(1.2)); draw(Y--P, linewidth(1.2)); draw(circle((-0.5487676153490452,0.23096704811575783), 0.5953920330094988), linewidth(1.2)); draw(P--X, linewidth(1.2)); /* dots and labels */ dot(A,linewidth(4.pt) + qqqqcc); label("$A$", (-0.641252415982058,0.8572858397873546), NW * labelscalefactor,qqqqcc); dot(D,linewidth(4.pt) + qqqqcc); label("$D$", (-1.0720851531544535,-0.23246755423693838), SW * labelscalefactor,qqqqcc); dot(B,linewidth(4.pt) + qqqqcc); label("$B$", (0.6222822501790012,0.8500449534483228), NE * labelscalefactor,qqqqcc); dot(C,linewidth(4.pt) + qqqqcc); label("$C$", (1.0205309988257534,-0.16005869084661992), NE * labelscalefactor,qqqqcc); dot(P,linewidth(4.pt) + qqqqcc); label("$P$", (-0.051120179350961574,0.35404423922464123), SW * labelscalefactor,qqqqcc); dot(X,linewidth(4.pt) + qqqqcc); label("$X$", (0.21317217202370123,1.689987768776017), NE ,qqqqcc); dot(Y,linewidth(4.pt) + qqqqcc); label("$Y$", (-1.2386255389521863,0.2418105009696476), NE * labelscalefactor,qqqqcc); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] since $APXB$ cyclic and $YD||XC$ it is trivial that $APDY$ is cyclic. the rest is to show $PY=PC$ so it is enough to show $\triangle XPY=\triangle XPC$ it is easy since we have $\angle AYP=\angle ADP=\angle PCB$ and $\angle PXB=\angle PXA$ also $XP=XP$! finished.

YD || XC so ∠AYD = 180 - ∠AXB = ∠APD ---> YAPD is cyclic. ∠YDA = ∠YPA so we have to prove PC = PY. we know PC = PD so instead we can prove PY = PD. ∠PYD = ∠PAD and ∠PDY = ∠PAX = ∠PBC = ∠PAD ---> ∠PYD = ∠PDY ---> PD = PY.

Let $R=AD\cap BC$. Let $\angle PAX=\angle PBX=\alpha$ and $\angle XAB=\beta$. So we get $\angle PBA=\alpha+\beta$ $\implies$ $\angle DAP=\alpha$ $\implies$ $\angle RDC=2\alpha+\beta$. $\angle CDY=\angle DCR=\angle RDC=2\alpha +\beta$. So we have that $DC$ is external angle bisector of $\angle ADY$ and $AC$ is internal angle bisector of $\angle DAY$. So $C$ is $A-\text{excenter}$ of $\triangle DAY$ and the result follows immediately.

Nice problem. We first make the following important observation. Claim : Quadrilateral $AYDP$ is cyclic. Proof : Simply note that, \[\measuredangle AYD = \measuredangle XYD = \measuredangle YXC = \measuredangle AXB = \measuredangle APD\]Thus, it is clear that $AYDP$ is indeed cyclic. Now, comes the key claim. Claim : $\triangle XPY \cong \triangle XPC$ Proof : We notice that, \[\measuredangle XYP = \measuredangle AYP = \measuredangle ADP = \measuredangle PCB = \measuredangle PCY\]and \[\measuredangle PXY = \measuredangle PXA = \measuredangle PBA = \measuredangle BAP = \measuredangle BXP = \measuredangle CXP\]Further, $\triangle XPY$ and $\triangle XPC$ share a side - $XP$ which implies that they are indeed congruent as desired. In particular, this implies that $PY=PC$. Now, it is a simple angle chase. Note that, \[\measuredangle YDA = \measuredangle YPA = \measuredangle YCP + \measuredangle PCY = 2\measuredangle YCP = 2\measuredangle YCA\]as was needed.

AlastorMoody wrote: Let $YP \cap BC=X$ X is already defined as $ \odot APB \cap BC$