Assume that $\angle BAP=\angle CAD$ $\frac{AD}{\sin\angle ACD}=\frac{CD}{\sin\angle DAC}$ $~$ $~$,$~$ $~$ $\frac{CD}{\sin\angle DFC}=\frac{DF}{\sin\angle DCF}$ $\Longrightarrow$ $\frac{AD}{DF}=\frac{\sin\angle ACD\cdot\sin\angle DFC}{\sin\angle DAC\cdot\sin\angle DCF}$ Likewise $~$ $\frac{AB}{BE}=\frac{\sin\angle AEP\cdot\sin\angle APB}{\sin\angle BAP\cdot\sin\angle BPE}$ $\Longrightarrow$ $\frac{AD}{DF}=\frac{AB}{BE}$ $~$ $~$ $\therefore BD//EF$ $~$ $~$ ($P$ is Miquel point)

Assume that $BD//EF$ Like the above, we can get $~$ $ \frac{AD}{DF}=\frac{\sin\angle ACD\cdot\sin\angle DFC}{\sin\angle DAC\cdot\sin\angle DCF} $ $~$ and $~$ $ \frac{AB}{BE}=\frac{\sin\angle AEP\cdot\sin\angle APB}{\sin\angle BAP\cdot\sin\angle BPE} $ $\therefore \frac{\sin\angle ACD}{\sin\angle DAC}=\frac{\sin\angle AEP}{\sin\angle BAP}$ $\Longrightarrow$ $\frac{AD}{DC}=\frac{AP}{EP}$ $~$ $~$ $\therefore\triangle DAC\sim\triangle PAE$ $~$ (SAS) $\therefore\angle PAB=\angle DAC$

See here. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1484879&sid=99d8be5eff1c9f23b1e8f0b44cf88326#p1484879

My solution: Lemma: Let $ ABCD $ be a convex quadrilateral . Let $ E=AB \cap CD, F=AD \cap BC $ and $ M $ be the Miquel point of $ ABCD $ . Let $ T $ be a point satisfy $ TB \parallel CD $ and $ TD \parallel CB $ . Then $ AT, AM $ are isogonal conjugate of $ \angle BAD $ Proof of the lemma : Let $ W, X, Y, Z $ be the midpoint of $ BF, AC, BD, EF $, respectively . From Newton line theorem we get $ X, Y, Z $ are collinear . Since $ \angle YWZ=180^{\circ}-\angle BAD=\angle DME $ , so from $ \frac{MD}{ME}=\frac{DF}{EB}=\frac{WY}{WZ} $ (Notice $ \triangle MDF \sim \triangle MEB $ ) we get $ \triangle WYZ \sim \triangle MDE $ , hence combine with $ AT \parallel XYZ $ we get $ \angle TAD=\angle ZYW=\angle EDM=\angle BAM $ . Back to the main problem : Let $ T $ be the point satisfy $ BTDC $ is a parallelogram . From lemma we get $ \angle BAP=\angle CAD \Longleftrightarrow A, T, C $ are collinear $ \Longleftrightarrow \frac {AB}{AE}=\frac{AT}{AC}=\frac{AD}{AF} \Longleftrightarrow BD \parallel EF $ Q.E.D

We apply Ceva's theorem to $\triangle ABD$ and point $P$. We obtain $$\frac{\sin\angle BAP}{\sin\angle PAD}=\frac{\sin\angle PDB}{\sin\angle PBD}\cdot\frac{\sin\angle PBA}{\sin\angle PDA}=\frac{PB}{PD}\cdot\frac{\sin\angle EBP}{\sin\angle PDF} $$Since $P$ is the center of spiral similarity sending $EB$ to $DF$, we further obtain $$\frac{\sin\angle BAP}{\sin\angle PAD}=\frac{PB}{PD}\cdot\frac{\sin\angle EBP}{\sin\angle PDF}=\frac{PB}{PD}\cdot\frac{\sin\angle EBP}{\sin\angle BEP}=\frac{PB}{PD}\cdot\frac{EP}{BP}=\frac{PE}{PD}=\frac{EB}{DF}$$where the last equality follows from $\triangle PEB\sim\triangle PDF$. On the other hand, applying Ceva's theorem to $\triangle ABD$ and point $C$, we obtain $$\frac{\sin\angle DAC}{\sin\angle BAC}=\frac{\sin\angle CBD}{\sin\angle BDC}\cdot\frac{\sin\angle CDA}{\sin\angle ABC}=\frac{\sin\angle EDF}{\sin\angle EBF}\cdot\frac{\sin\angle EFB}{\sin\angle FED}\cdot\frac{CD}{BC}\cdot\frac{CF}{CE}=\frac{EB}{DF}\cdot\frac{CD}{BC}\cdot\frac{CF}{CE}$$ Therefore \begin{align*} &\angle BAP=\angle CAD\\ \iff &\frac{\sin\angle BAP}{\sin\angle PAD}=\frac{\sin\angle DAC}{\sin\angle BAC}\\ \iff &\frac{CD}{BC}\cdot\frac{CF}{CE}=1\\ \iff & BD\|EF \end{align*}

spiral similarity is key of the solution. firstly let us proof problem for case $BD\parallel EF$. with proof with contradiction you can prove for other cases we haven't $\angle BAP=\angle CAD$ PROOF FOR CASE $BD\parallel EF$. note that $AFPB,AEPD$ are cyclic (with little angle chasing) . So we need to prove that $\angle BAP=\angle BFP =\angle CAD$ in other word : $\triangle CPF \sim \triangle CDA , \triangle CPE \sim \triangle CBA $. define$P'$ Such : $\triangle CP'F \sim \triangle CDA , \triangle CP'E \sim \triangle CBA $ (we have spiral similarity) easily we can prove exist $P'$ And we know that: $P' \equiv P$ . and we are done.

Observe $P$ is the Miquel Point of $ABCD,$ and quadrilaterals $AEPD$ and $AFPB$ are cyclic. Suppose $BD\parallel EF.$ Then $$\frac{BC}{CF} = \frac{BD}{EF} = \frac{AD}{AF}.$$But since $\frac{PD}{PC} = \frac{AD}{BC}$ by spiral similarity properties, $$\frac{PD}{PC} = \frac{AD}{BC} = \frac{AF}{CF}.$$Since $\angle AFC = \angle DPC,$ $\triangle AFC \sim \triangle DPC \implies \angle CAD = \angle CDP = \angle BAP$ as desired. $\square$ Suppose $\angle BAP=\angle CAD.$ Then $$\triangle ACD \sim \triangle AEP \implies \frac{AD}{AC} = \frac{AP}{AE},$$and $$\triangle ACB \sim \triangle AFP \implies \frac{AB}{AC} = \frac{AP}{AF}.$$Dividing the two equations finishes. $\square$

If $BD\parallel EF$, consider an inversion $\Phi $ with center $A$ and radius $\sqrt{AB\cdot AF}$, and the reflection $\Gamma $ wrt the angle bisector of $\angle BAD$ . Set $\Theta:=\Phi\circ \Gamma$. Note that $\Theta(B)=F$, $\Theta(D)=E$, thus $\Theta(DE)=(ABF)$, $\Theta(BF)=(ADE)$, therefore $$\Theta(C)=\Theta(BF\cap DE)=(ADE)\cap (ABF)$$which is the Miquel point of $ABCDEF$, i.e. $P$. Hence $\Theta$ swaps $C$ and $P$, meaning that $\angle BAC=\angle PAD$.