$P$ 是 $\triangle A B C$ 的内心, $B C=4, \angle B A C=90^{\circ}$, 则 $\triangle P B C$ 的外接圆半径为
$\because P$ 是 $\triangle A B C$ 的内心

$\therefore A D$ 平分 $\angle B A C, B P$ 平分 $\angle A B C$,

$\therefore \angle A B P=\angle C B P, \angle B A P=\angle C A D$,

$\therefore \widehat{B D}=\widehat{C D}$

$\therefore B D=C D$

$\therefore \angle D B C=\angle C A D$,

$\therefore \angle D B C=\angle B A P$,

$\because \angle D B P=\angle C B P+\angle D B C, \angle D P B=\angle A B P+\angle B A P$,

$\therefore \angle D B P=\angle D P B$,

$\therefore D P=D B$

$$
\therefore B D=D P=C D
$$

$\therefore D$ 为 $\triangle P B C$ 的外接圆的圆心, $B D$ 为 $\triangle P B C$ 的外接圆半径

$\because \angle B A C=90^{\circ}$,

$\therefore B C$ 是直径,

$\therefore \angle B D C=90^{\circ}$

$$
\begin{gathered}
\therefore 2 B D^{2}=B C^{2}=16 \\
\therefore B D^{2}=8
\end{gathered}
$$

$$
B D=2 \sqrt{2}
$$

$\therefore \triangle P B C$ 外接圆的半径为 $2 \sqrt{2}$