12. B. As shown in Figure 5, given that the side length of square $A B C D$ is $1$, and $P, Q$ are two points inside it, and
$$
\angle P A Q=\angle P C Q=45^{\circ} \text {. }
$$

Find the value of $S_{\triangle P A B}+S_{\triangle P C Q}+S_{\triangle Q A D}$.
12. B. As shown in Figure 10, rotate $\triangle A Q D$ 90° clockwise around point $A$ to $\triangle A Q^{\prime} B$, and rotate $\triangle C Q D$ 90° counterclockwise around point $C$ to $\triangle C Q^{\prime \prime} B$. Connect $P Q^{\prime}$ and $P Q^{\prime \prime}$. Then
$$
\begin{array}{l}
\triangle A P Q^{\prime} \\
\cong \triangle A P Q, \\
\triangle C P Q^{\prime \prime} \cong \triangle C P Q . \\
\text { By } \angle A B Q^{\prime}+\angle C B Q^{\prime \prime} \\
=\angle A D Q+\angle C D Q=90^{\circ},
\end{array}
$$

we know that points $Q^{\prime}$, $B$, and $Q^{\prime \prime}$ are collinear, and $B Q^{\prime}=D Q=B Q^{\prime \prime}$.
$$
\begin{array}{l}
\text { Then } S_{\triangle P B Q^{\prime}}=S_{\triangle P B Q^{\prime \prime}} \text {. } \\
\text { Therefore, } S_{\triangle P A B}+S_{\triangle P C Q}+S_{\triangle Q A D} \\
=S_{\triangle P A Q}+S_{\triangle P B C}+S_{\triangle Q C D} \\
=\frac{1}{2} S_{\text {square }ABCD}=\frac{1}{2} .
\end{array}
$$