[asy][asy] draw((0,4)--(8,10)); draw((4,-2)--(0,4)); draw((4,-2)--(5,-1)); draw((5,-1)--(8,4)); draw((8,4)--(8,10)); label("A",(8,10),NE); label("B",(0,4),W); label("C",(4,-2),S); label("D",(5,-1),E); label("E",(8,4),E); draw((0,4)--(8,4),dashed); draw((0,4)--(5,-1),dashed); draw(rightanglemark((8,10),(8,4),(0,4))); draw(rightanglemark((0,4),(5,-1),(4,-2)));[/asy][/asy] Here is a visual. I am unable to solve the problem at the moment. Can anybody understand the problem and find a solution?

Let $F = CD \cap AE.$ Then, notice that since $\angle FCA = \angle BCA, \angle FAC = \angle BAC$, we have that $F$ is the reflection of $B$ over $AC$. Let $BF \cap AC = G.$ Then, since $\angle BGC = \angle BDC = 90$, we've that $BGDC$ is cyclic and so hence $FG \cdot FB = FD \cdot FC.$ Analogously, we also have that $FG \cdot FB = FE \cdot FA,$ and so hence $FD \cdot FC = FE \cdot FA \Rightarrow ACDE$ cyclic, solving (b). As for (a), observe that since $\angle AGB = 90$, we have that $G = P.$ Analogously, $G \in (\triangle BCD).$ Therefore, since $\angle PCB = \angle PCF = \angle PCD$ we have that $P$ is the midpoint of arc $BD$ on $(\triangle BCD)$, which gives that $BP = PD.$ Analogously, we have that $BP = PE$, hence implying part (a). $\square$

Clearly $P$ is the projection from $B$ to $AC$ thus quadrilaterals $ABPE$ and $CBPD$ are cyclic. This implies that $PB=PD=PE$, completing (a). For (b) we just observe that $$\angle CAE + \angle CDE = \angle PBE + (90^{\circ} + \angle BDE) = 180^{\circ}.$$

$\angle{AEB}=\angle{APB}=90^o=\angle{BPC}=\angle{BDC}$ Implies $BPDC$ cyclic. $\angle{EAP}=\angle{BAP}\Rightarrow EP=BP$ Similarly $\angle{DCP}=\angle{BCP}\Rightarrow DP=BP$ So, $P$ is the circumcenter of $BDE$ Again, $\angle{AED}=\angle{AEB}+\angle{BED}=90^o+\frac{1}{2}\angle{BPD}=90^o+\frac{1}{2}(180^o-\angle{DCB})=180^o-\angle{DCA}$

An intuitive way to add point $P$ for part b) if we don't have part a) as a hint is to consider $AE\cap CD=X$. Note that $\triangle BAC\cong\triangle XAC$. So $BX\perp AC$ at, say, point $P$. We finish by noticing that $$\measuredangle CDE=\measuredangle XDE=\measuredangle XBE=\measuredangle PBE=\measuredangle PAE=\measuredangle CAE.$$