Let $O_1$, $O_2$ the circumcenters of $\triangle ABC$,$\triangle ADE$ respectively, and let $X=EC\cap \odot (ADE)$, $Y=BD\cap \odot (ABC)$. Since $DXAE$ is cyclic we get $\measuredangle AXE=\measuredangle ADE=\measuredangle CAP$ $\Longrightarrow$ $\measuredangle XAP=\measuredangle PCA$ similarly we get $\measuredangle YAP=\measuredangle PDA$, so $$\measuredangle BCP=\measuredangle PCA+\measuredangle BCA=\measuredangle DAP+\measuredangle XAP=\measuredangle DAX=\measuredangle DEP $$hence $BC\parallel DE$ $\Longrightarrow$ $\triangle PCB\sim \triangle EPD...(1)$. So from $\measuredangle XAP=\measuredangle PCA$ and $\measuredangle YAP=\measuredangle PDA$ we get $PX.PC=AP^2=PY.PD$ $\Longrightarrow$ $XYDC$ is cyclic, by simple angle-chasing we get $$\measuredangle DAE=\measuredangle DXE=\measuredangle DYB=\measuredangle CAB$$$\Longrightarrow$ $\measuredangle CO_1B=2\measuredangle CAB=2\measuredangle DAE= \measuredangle DO_2E$ $\Longrightarrow$ $\triangle CO_1B\sim \triangle EO_2D...(2)$. Finally, by $(1)$ and $(2)$ we get $BO_1PC$ and $DO_2PE$ are similar $\Longrightarrow$ $\measuredangle BPO_1=\measuredangle DPO_2$, hence $O_1$, $O_2$ and $P$ are collinear.

We will show that there exists a dilation $\sigma$ with center $P$ which maps $\Omega_1=\odot(ABC)$ onto $\Omega_2=\odot(ADE)$. Thus, $\sigma$ will map $O_1 \mapsto O_2$ ($O_i :=$ center of $\Omega_i$) and $P, O_1, O_2$ are collinear. First note that $BC\parallel ED$ since $\angle BCA+\angle ADE=\angle ADC$. Alternatively one could argue with the isogonal conjugate line to $AP$ in $ACD$ (i.e. a line $AQ$ with $\angle CAP=\angle QAD$ and $\angle PAD=\angle CAQ$) which is obviously parallel to both $BC$ and $ED$. Because of $BC\parallel ED$, there exists a dilation $\sigma$ with $P\mapsto P$, $B\mapsto D$, and $C\mapsto E$. Let $A'=A\sigma$ denote the image of $A$ $\Longrightarrow P,A,A'$ collinear. Then $\triangle ABC$ and $\triangle A\sigma B\sigma C\sigma = \triangle A'DE$ are similar triangles, and thus $\angle ACB=\angle A'ED$. But also $\angle ACB=\angle PAD=\angle A'AD$ (from the problem statement). $\Longrightarrow$ $\angle DEA'=\angle DAA'$ $\Longrightarrow$ $D,E,A,A'$ cyclic. We are done because $\sigma$ maps $\Omega_1=\odot(ABC)$ onto $\Omega_3=\odot(A'DE)$ but we have just shown that $\Omega_3=\Omega_2$.

Denote the circumcenters of $\triangle ABC, \triangle ADE$ by $O_1, O_2$, respectively. Let $l$ be the line isogonal to $AP$ wrt $\angle CAD$. Then from $\angle PAD = \angle ACB$ and $\angle CAP = \angle EDA$ it follows that the lines $BC$ and $DE$ are both parallel to $l$. Let $\infty_{l}$ be the point at infinity on the line $l$. By this isogonal line lemma applied to the pairs of isogonal lines $(AC, AD)$ and $(AP, A\infty_{l})$ we get that $AB$ and $AE$ are isogonal wrt $\angle CAD$. Now we have $\angle BO_1C = 2\angle BAC$ and $\angle DO_2E = 2\angle DAE$, hence $\angle BO_1C = \angle DO_2E$. Since the triangles $\triangle BO_1C$ and $\triangle DO_2E$ are both isosceles and $BC \parallel DE$, it follows that they are homothetic. Obviously $BD \cap CE = P$ is the center of the homothety mapping $\triangle BO_1C$ to $\triangle DO_2E$, hence $O_1, O_2$ and $P$ are collinear.