1. Square $A B C D$ has a side length of $5, E$ is a point on side $B C$ such that $B E=3, P$ is a point on the diagonal $B D$ such that $P E+P C$ is minimized. Then $P B=$ $\qquad$
= $1 \cdot \frac{15 \sqrt{2}}{8}$.
Since $P E+P C=P E+P A$, therefore, when $A$, $P$, and $E$ are collinear, $P E+P A$ is minimized.

As shown in Figure 7, establish a rectangular coordinate system, with $B$ as the origin and $B A$ as the $x$-axis. Then
$$
\begin{array}{l}
l_{B D}: y=x, \\
l_{A E}: 3 x+5 y=15 .
\end{array}
$$

Therefore, $P\left(\frac{15}{8}, \frac{15}{8}\right)$.
Thus, $P B=\frac{15 \sqrt{2}}{8}$.