6. As shown in the figure, in the convex quadrilateral $A B C D$, $A B=2$, $P$ is the midpoint of side $A B$. If $\angle D A B=\angle A B C$ $=\angle P D C=90^{\circ}$, then the minimum possible value of the area of quadrilateral $A B C D$ is ( ).
(A) 4
(B) $3 \sqrt{2}$
(C) $\frac{9}{2}$
(D) $2+2 \sqrt{21}$
6. (A).

If the garden is set as $A D=x, B C$ $=y$, then by the given information we get $P D=\sqrt{1+x^{2}}, \quad P C=$ $\sqrt{1+y^{2}}, C D=\sqrt{y^{2}-x^{2}}$ (using the Pythagorean theorem).

Construct $D Q \perp B C, Q$ as the foot of the perpendicular, then $C Q=y-x, D Q=2$. Thus,
$$
y^{2}-x^{2}=2^{2}+(y-x)^{2},
$$

which simplifies to $x^{2}-x y+2=0$.
Let $S$ be the area of $A B C D$, then $S=\frac{(x+y) \cdot 2}{2}=x+$ $y$. Substituting $y=S-x$ into (1) gives $2 x^{2}+x S+2=0$. Let the discriminant $\Delta=S^{2}-4 \times 2 \times 2 \geqslant 0$, solving this yields $S \geqslant 4$.

Furthermore, when $x=1, y=3$, $S=4$, hence the minimum area of $A B C D$ is 4.