Define $E=AB\cap CD$, $F=AC\cap BD$, $\omega$ is the circumcircle of $ABCD$, $O$ is the center of $\omega$, $R$ is the length of radius of $\omega$, and $\ell$ is the internal bisector of $\angle AED$. $\textbf{Lemma :}$ There exist a circle $\Gamma$ which is orthogonal to $\omega$ and tangent to both $AB$ and $CD$. $\textit{Proof :}$ Note that it suffices to prove that there exists point $Y$ on $\ell$ such that $YO^{2}=YE^2\cdot (\sin \frac{\angle AED}{2})^2 + R^2$. To do this we'll use cartesian coordinate. WLOG $E=(0,0)$, $\ell$ is line $y=x$, and $O=(p,q)$. For any point $K=(k,k)$ on line $\ell$, we consider function $f(K)=(k-p)^2+(k-q)^2-2k^{2}\cdot (\sin \frac{\angle AED}{2})^2-R^2$. It's easy to see that $f(K)$ will become a continuous function (it is a polynomial). Furthermore we also have when $K=E$, $f(K) > 0$ and when $K\in \omega$, $f(K) < 0$. Thus by Intermediate Value Theorem, there exist $K$ such that $f(K)=0$. This implies $KO^{2}=KE^{2}\cdot (\sin \frac{\angle AED}{2})^2 + R^2$ and the lemma is now proved. Now define $\Gamma$ as a circle which is orthogonal to $\omega$ and tangent to both $AB$ and $CD$. We'll prove that this circle is also tangent to both $(ABP)$ and $(CDP)$ which will finish this problem. By symmetry it suffices to prove $\Gamma$ is tangent to $(ABP)$. Note that $\angle PBC=\angle PDA$ and $\angle PCB=\angle PAD$ implies that $P$ is the center of spiral similarity which sends $DA$ to $BC$. Thus, we can get $ADPF$ and $BPCF$ are both cyclic. After that, by simple angle chasing we can get $ABPO$ and $CDPO$ are both cyclic. Define $M$ as the contact point of $\Gamma$ and $AB$, $N$ as the intersection of segment $MO$ with $\omega$ and $N'$ as the intersection of line $MO$ with $\Gamma$ where $N'\neq M$ Notice that inversion around $\Gamma$ sends $N$ to $M$, hence $ON\times OM=R^{2}$. But as $\Gamma$ is orthogonal to $\omega$ we also have $ON'\times OM=R^{2}$, so we must have $N=N'$. After that, to prove that $\Gamma$ is tangent to $(ABP)$ it suffices to prove that tangent of $\Gamma$ at $M$ is parallel to tangent $(ABP)$ at $O$ which is true. (because line $AB$ is line tangent to $(ABP)$ which pass through $M$ and $OA=OB$ implies tangent to $(ABP)$ at $O$ is parallel to $AB$) Therefore, $\Gamma$ is tangent to $(ABP)$ and the problem is now solved.