Consider the negative inversion at $P$ that swaps $A,C$ and $B,D$. The line $AD$ inverts to $\omega$, the circumcircle of $\triangle BCP$. Points $A_1$ and $D_1$ are sent to $A'$, the second intersection of $A_1P$ with $\omega$, and $D'$, the second intersection of $D_1P$ with $\omega$. [asy][asy] size(10cm); pair A = dir(215), D = dir(-35), B = dir(130), C = dir(80), P = extension(A, C, B, D), A1 = extension(A, D, P, rotate(90, P)*A), D1 = extension(A, D, P, rotate(-90, P)*D), O = circumcenter(B, P, C), Dp = 2*O-B, Ap = 2*O-C; draw(unitcircle, heavyred); draw(A--B--C--D^^D1--A1, red); draw(circle(O, abs(O-B)), orange); draw(arc(circumcenter(D1, P, A1), circumradius(D1, P, A1), 0, 180), fuchsia); draw(D1--Dp^^A1--Ap, heavymagenta); string[] names = {"$A$", "$B$", "$C$", "$D$", "$P$", "$A_1$", "$D_1$", "$A'$", "$D'$"}; pair[] points = {A, B, C, D, P, A1, D1, Ap, Dp}; pair[] ll = {dir(240), dir(150), dir(60), dir(300), dir(270), A1, D1, dir(40), dir(40)}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] It is equivalent to show that line $A'D'$ is parallel to $BC$. But due to the right angles, $A'$ and $D'$ are simply the antipodes of $C$ and $B$ in $\omega$, respectively. Therefore $BCD'A'$ is a rectangle.

Actually, a simple angle chase does the job as well: We only need to prove that $\angle CBD+\angle DPA_1=\angle PD_1A_1$. But \[\angle CBD+\angle DPA_1=\angle CAD+\angle DPA_1=90^\circ-\angle AA_1P+\angle DPA_1=90^\circ-\angle ADP=\angle PD_1A_1.\]

We extend $PD_1$ and $PA_1$ to meet $BC$ at $B_1$ and $C_1$ respectively. Note that if we show $A_1B_1C_1D_1$ is cyclic then we are done Simple angles show that the pairs $(A_1,D_1),(B_1,C_1)$ are isogonal. So $\angle APD_1=\angle A_1PD=\angle BPC_1$ Also $\angle PAD_1=\angle PBC_1$ Thus we have $\angle AD_1P=\angle BC_1P$(using triangle property that A+B+C=180) Thus $A_1B_1C_1D_1$ is cyclic