Let $A B C D$ be a square, and point $P$ lies on line $C D$, with the distance from point $P$ to point $A$ being 1. Then the maximum possible value of the area of quadrilateral $P^{\prime} A B C$ is
(A) $\frac{\sqrt{5}+2}{4}$
(B) $\sqrt{2}$
(C) $\frac{\sqrt{5}+1}{2}$
(D) $\sqrt{5}-1$
3. (A).

As shown in the figure, let $\angle P A D=\theta$, then $P D=\sin \theta, A D=\cos \theta$, hence
$$
\begin{aligned}
S_{P A B C} & =\cos ^{2} \theta+\frac{1}{2} \sin \theta \cos \theta \\
& =\frac{1+\cos 2 \theta}{2}+\frac{1}{4} \sin 2 \theta \\
& =\frac{\sqrt{5}}{4} \sin (2 \theta+\alpha)+\frac{1}{2} \\
& \leqslant \frac{\sqrt{5}}{4}+\frac{1}{2},
\end{aligned}
$$

where $\alpha=\arctan 2$. Equality holds when $\theta=\frac{\pi}{4}-\frac{1}{2} \arctan 2$.