Denote by $ B',C'$ the symmetrical points of $ B,C$, respectively, through the line $ OP$. Denote by M the intersection of $ OP$ with the circle.$ MO$ intersects the second time the circle $ M'$. $ \angle PC'B=\frac {\widehat {BC'}} {2}=\frac {\widehat {B'C}} {2}=\frac {\widehat {CM}-\widehat {MB}} {2}=\frac {\widehat {DM}-\widehat {MB}} {2}=\angle BEM \Rightarrow$ the quadrilateral $ BEPC'$ is cyclic $ \Rightarrow \angle PBE=\angle PC'E=\angle PCE$ But $ \angle PBE=\angle DBA=\angle DCA$ So, $ \angle PCE=\angle DCA \Rightarrow EC \perp CA$

Dear Mathlinkers, a synthetic proof: let F the point of intersection of CE , (O) and Tc the tangent to (O) at C and O' the point of intersection of AF and CD. According to Pascal's theorem aplied to CDBAFCTc, O', E and P are collinear. It follow that O'=O. ... Sincerely Jean-Louis

Denote by N the intersection point of the lies $ BD$ and $ PC$. W.L.O.G. assume $ R=1$, where $ R$ is the radii of circle $ (O)$. Denote also $ \angle{ABC}=x$ and $ \angle{CAB}=y$. As $ CD$ passes on $ O$, from right triangle $ ABD$, $ DB=2cos{y}$. From $ CDN$ we get $ CN=2tg{y}$, $ DN=\frac{2}{cos{y}}$. $ BN=\frac{CN^{2}}{DN}=\frac{2sin^{2}{y}}{cos{y}}$. Note that $ \angle{BPN}=\angle{BPC}=x-y$. By sine law applied in triangle $ BNP$, $ NP=\frac{2sin^{2}{y} cos{x}}{cos{y} sin{(x-y)}}$. So, $ CP=NP+PC=\frac{2sinxsiny}{sin(x-y)}$. By Menelaus theorem in triangle $ COP$ by controversial $ DEN$, $ \frac{CO}{OD} \cdot {\frac{DE}{EN}} \cdot {\frac{NP}{PC}}=1$, but $ CO=OD$, $ \frac{NP}{PC}=tgy\cdot{ctgx}$. It follows that $ \frac{DE}{EN}=tgx\cdot{ctgy}$, but from triangle $ DCN$, $ \frac{DE}{EN}=\frac{[DCE]}{[NCE]}=\frac{DC}{CN} \cdot {\frac{sin{\angle DCE}}{sin{\angle NCE}}=ctgy \cdot {tg{\angle{DCE}}}}$. It means $ ctgy \cdot {tg{\angle{DCE}}=tgx\cdot{ctgy}}$, which yields $ \angle{DCE}=x$. Finally, $ \angle{DCE}=x=\angle{ADC}$, which immediately follow $ AD||CE$, from which $ \angle{ACE}=180-\angle{DAC}=90$, as desired.