5. Let $P$ be a point inside square $A B C D$, with $P A=1$, $P B=2, P C=3$. Then the area of $\triangle P B C$ is ( ).
(A) $2+\frac{\sqrt{2}}{2}$
(B) $2-\frac{\sqrt{2}}{2}$
(C) $2+\sqrt{2}$
(D) $2-\sqrt{2}$
5. A.

As shown in Figure 3, $\triangle A P B$ is rotated $90^{\circ}$ clockwise around point $B$ to get $\triangle C Q B$.
Connect $P Q$.
Since $\angle P B Q=90^{\circ}$,
$$
P B=Q B=2 \text{, }
$$

then $\angle P Q B$
$$
\begin{array}{l}
=\angle Q P B=45^{\circ}, \\
P Q=2 \sqrt{2} .
\end{array}
$$

In $\triangle P Q C$,
$$
P C=3, C Q=1, P Q=2 \sqrt{2} \text{. }
$$

By $P C^{2}=C Q^{2}+P Q^{2} \Rightarrow \angle P Q C=90^{\circ}$.
Then $S_{\text {quadrilateral } P B Q C}=S_{\triangle P B Q}+S_{\triangle P Q C}$
$$
=\frac{1}{2} \times 2 \times 2+\frac{1}{2} \times 2 \sqrt{2} \times 1=2+\sqrt{2} \text{. }
$$

Since $\angle P Q B=45^{\circ}$,
$$
\angle B Q C=45^{\circ}+90^{\circ}=135^{\circ} \text{. }
$$

Draw $C H \perp B Q$ at point $H$. Then
$$
\angle C Q H=45^{\circ} \text{. }
$$

Therefore, $C H=\frac{\sqrt{2}}{2}$,
$S_{\triangle B Q C}=\frac{1}{2} B Q \cdot C H=\frac{1}{2} \times 2 \times \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$.
Thus, $S_{\triangle P B C}=S_{\text {quadrilateral } P B Q C}-S_{\triangle B Q C}$
$$
=2+\sqrt{2}-\frac{\sqrt{2}}{2}=2+\frac{\sqrt{2}}{2} \text{. }
$$