<Qua-2025-757>, hint

Let <DAP=x, <CBP=y . Since BC=BP, <DAB=60+x, BCD=180-(60+x)=120-x

<BPC=180-(120-x)-y=60+x-y=90-y/2. Hence 60=2x-y, 

Similarly <APD=60+y-x. Thus <BPC+<BPA+<APD=180

. <APD=60+y-x=60+(2x-60)-x=x=<DAP. Triangles ADP and ABP are isosceles' 

Hence  BD is angle bisectors of <ADP and <ABP, <ABD=30

AE=PD ( DAP is isosceles ) PD/sin30=2r (r is radius of circumcircle of ABCD)

PD=r.