Edited, I had a typo making the problem wrong.

Let's bash! Inversion with center at $A$ and arbitrary radius. Let $B, C, D, P\mapsto E, F, G, H$ respectively. Let $\angle{GAH}=a$ and $\angle{HAF}=b$. Note that $AFHG$ is a parallelogram and $$\angle{EAF}=\angle{BAC}=t, \angle{BPC}=\angle{APC}-\angle{APB}=\angle{AFH}-\angle{AEH}=2t.$$So, $\angle{AEH}=(180^{\circ}-a-b)-2t\implies \angle{EHF}=t$. Let $\angle{ABD}=\angle{AGE}=x$ and $\angle{BCA}=\angle{AEF}=y$. Note that $\angle{PBD}=|(a+t)-x|$ and $|\angle{BCA}-\angle{PCA}|=|\angle{AEF}-\angle{AHF}|=|y-a|$. So, it's enough to prove that $x=y+t$. By Ceva's, we've $$\frac{\sin (x)}{\sin (180^{\circ}-x-a-b)}\times \frac{\sin (a+b+t)}{\sin (a+t)}\times \frac{\sin (b+t)}{\sin (a+b+t)}=1$$and $$\frac{\sin (y)}{\sin (180^{\circ}-y-a-b-2t)}\times \frac{\sin (t)}{\sin (a)}\times \frac{\sin (b)}{\sin (t)}=1.$$ We have \begin{align*} & \ \frac{\sin (y+t)}{\sin (180^{\circ}-(y+t)-a-b)}\times \frac{\sin (a+b+t)}{\sin (a+t)}\times \frac{\sin (b+t)}{\sin (a+b+t)}=1\\ & \iff \sin (y+t) \sin (b+t)=\sin (a+b+y+t)\sin (a+t) \\ & \iff \sin (y)\left( \sin (t+b)\cos (t)-\sin (t+a)\cos (a+b+t)\right) =\cos (y)\left( \sin (a+b+t)\sin (t+a)-\sin (t)\sin (t+b)\right)\\ & \iff \sin (y)\left( \frac{(\sin (2t+b)+\sin (b))-(\sin (2t+2a+b)+\sin (-b))}{2} \right) =\cos (y)\left( \frac{(\cos (b)-\cos (b+2a+2t))-(\cos (b)-\cos (2t+b))}{2}\right)\\ & \iff \sin (y)\left( \sin (b)-\frac{\sin (2t+2a+b)-\sin (2t+b)}{2}\right) =\cos (y)\left( -\sin (2t+a+b)\sin (-a)\right)\\ & \iff \sin(y)\left( \sin (b)-\cos (a+b+2t)\sin (a)\right) =\cos (y)\left( \sin (a)\sin (a+b+2t)\right), \end{align*}which is true since $\sin (y)\sin (b)=\sin (y+a+b+2t)\sin (a)$. So, $$\frac{\sin (y+t)}{\sin (180^{\circ}-(y+t)-a-b)} =\frac{\sin (x)}{\sin (180^{\circ}-x-a-b)}.$$This gives $$2\sin (y+t)\sin (x+a+b)=2\sin (x)\sin (y+t+a+b)\implies \cos (y+t-x-a-b)=\cos (x-y-t-a-b).$$Hence, we get $$\begin{rcases*} -2\sin(-a-b)\sin(y+t-x)=0 \\ a+b=\angle{DAC}\in (0,\pi ) \\ \begin{rcases*} y+t+a+2b=180^{\circ}\implies y+t\in (0,\pi ) \\ x=\angle{AGE}=\angle{ABC}<180^{\circ}\implies x\in (0,\pi ) \\ \end{rcases*} y+t-x\in (-\pi ,\pi) \\ \end{rcases*} y+t-x=0. \quad \blacksquare$$

Let $Q$ on $AC$ such that $QA=QB$ and $R\equiv \odot APD \cap PQ$. It's obvious that $BQPC$ is a cyclic quadrilateral. Now we have $QB^2=QA^2=QR\cdot QP\Rightarrow \overset{\triangle}{QBR}\sim \overset{\triangle}{QPB}\Rightarrow \angle QRB=\angle QBP=\angle ACP=\angle DAP=\angle DRP\Rightarrow B,R,D$ are collinear and so $\angle BCA=\angle BPQ=\angle QBR=\angle QBD=\angle QBP-\angle DBP=\angle ACP-\angle DBP$ and we are done.

Let $J,K$ be the circumcenters of $\triangle ACD , \triangle ABC$.$O$ lies on $AC$ s.t. $KO \perp AB$. so $O,P,KC,B$ are concyclic (by simple angle chasing).$BD$ cuts circle $(PBC)$ at $I$.It's obvious that $A,I,K$ are conllinear. so $\angle IAO=\angle KAI=\angle KCO=\angle AIO$ so $OI=AO=OB$ thus $O$ be the circumcenter of $\triangle ABI$ so $A$ be $C-excenter$ of $\triangle ICB$.$\angle PBD=\angle PCI=\angle PCA-\angle (ICA=ACB)$ so done.

Let $ T $ be the point such that $ \triangle BPT $ $ \stackrel{+}{\sim} $ $ \triangle APC $ $ \stackrel{+}{\sim} $ $ \triangle DPA $ and $ S $ be the intersection of $ AC, $ $ BT. $ Clearly, $ S $ lies on $ \odot (ABP), $ $ (CPT), $ so from $ \angle BPC $ $ = $ $ 2\angle BAC $ we get $ \angle BAC $ $ = $ $ \angle BTC $ $ \Longrightarrow $ $ T $ $ \in $ $ \odot (ABC), $ hence $$ \underbrace {\angle PBD = \angle PTA}_{ \because \ \triangle BPT \stackrel{+}{\sim} \triangle DPA}= \angle PTB - \angle ATB = \angle PCA - \angle ACB. $$

Attachments:

unless i'm mistaken the the statement should be: $\angle PBD= \angle BCA + \angle PCA $ with oriented angles . RH HAS

nice problem! Let K,S be on PB such that KP=PD,SP=PC. Since ABCS,ABDK are cyclic with easy angle chasing,and power of point satisfies to show PA^2=PK.PS or,PA^2=PD.PC.since P is A_Dumpty point in ADC we are done