3. In an isosceles right $\triangle ABC$, $AB=BC=5$, $P$ is a point inside $\triangle ABC$, and $PA=\sqrt{5}, PC=5$. Then $PB=\quad$ $\qquad$
3. $\sqrt{10}$.

As shown in Figure 3, draw $P E \perp$ $A B$ and $P F \perp B C$ intersecting $A B$ and $B C$ at points $E$ and $F$ respectively.
Let $P E=m$,
$$
P F=n \text{. }
$$

Using the Pythagorean theorem in $\triangle P A E$ and
$\triangle P C F$ respectively, we get
$$
\begin{array}{l}
m^{2}+(5-n)^{2}=5, \\
(5-m)^{2}+n^{2}=25 .
\end{array}
$$
Subtracting (1) from (2) gives $10(n-m)=20$, which simplifies to $m=n-2$.
Substituting into equation (1) yields $n^{2}-7 n+12=0$.
Solving this, we get $n_{1}=3, n_{2}=4$.
When $n=3$, $m=n-2=1$.
In the right triangle $\triangle P B E$, by the Pythagorean theorem,
$$
P B=\sqrt{m^{2}+n^{2}}=\sqrt{10} \text{. }
$$

When $n=4$, $m=n-2=2$, in this case, $P E > A E$, so point $P$ is outside $\triangle A B C$, which does not meet the problem's conditions, and thus is discarded.
Therefore, $P B=\sqrt{10}$.