8. (40 points) In $\triangle A B C$, it is given that $B C=A C$, $\angle B C A=90^{\circ}$, points $D$ and $E$ are on sides $A C$ and $A B$ respectively, such that $A D=A E$, and $2 C D=B E$. Let $P$ be the intersection of segment $B D$ and the angle bisector of $\angle C A B$. Find $\angle P C B$.

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The translation is provided as requested, maintaining the original formatting and structure.
8. Solution 1 As shown in Figure 14, let $F$ be the midpoint of side $BE$. Then $FC$ is parallel to $ED$.
Extend $ED$ to intersect the extension of $BC$ at point $Q$. Thus, $\triangle ABQ$ is an isosceles right triangle.
$$
\begin{aligned}
& \text{Then } \angle AQE \\
= & \angle ADE - 45^{\circ} \\
= & \angle AED - 45^{\circ} = \angle BQE.
\end{aligned}
$$

Therefore, $DQ$ bisects $\angle AQC$.
By symmetry, $BD$ bisects $\angle ABC$.
Thus, $P$ is the incenter of $\triangle ABC$.
Hence, $\angle PCB = 45^{\circ}$.
Solution 2 Let $AC = BC = a$,
$$
BF = FE = CD = x.
$$

Then $AD = AE = a - x$.
Thus, $\sqrt{2} a = AB = a - x + 2x = a + x$.
Therefore, $x = (\sqrt{2} - 1) a$, and $AD = (2 - \sqrt{2}) a$.
Thus, $\frac{AD}{CD} = \frac{(2 - \sqrt{2}) a}{(\sqrt{2} - 1) a} = \sqrt{2} = \frac{BA}{BC}$.
Therefore, $BD$ bisects $\angle ABC$.
Thus, $P$ is the incenter of $\triangle ABC$.
Hence, $\angle PCB = 45^{\circ}$.