2. Given a point $P$ outside the plane of a convex quadrilateral $A B C D$, and $E 、 F 、 G 、 H 、 M 、 N$ are the midpoints of $A B$ 、 $P C 、 A D$ 、 $B C$ 、 $E F$ 、 $G H$ respectively. Then ( ).
(A) $P 、 D 、 M 、 N$ are coplanar, and $P D=4 M N$
(B) $P 、 D 、 M 、 N$ are not coplanar, and $P D=$ $4 M N$
(C) $P 、 D 、 M 、 N$ are coplanar, and $P D \neq 4 M N$
(D) $P 、 D 、 M 、 N$ are not coplanar, and $P D \neq$ $4 M N$
2.A.

As shown in Figure 3, take the midpoint $Q$ of $CD$, and connect the midpoints of the sides of quadrilateral $ABCD$ in sequence to form parallelogram $GEHQ$, then connect $QE$.

By the property that the diagonals of a parallelogram bisect each other, we know that $QE$ passes through the midpoint $N$ of $GH$, and $N$ is the midpoint. Connect $FQ$.
By the Midline Theorem of a triangle, we have
$$
\overrightarrow{MN} = \frac{1}{2} \overrightarrow{FQ} = \frac{1}{4} \overrightarrow{PD} \text{.}
$$

Therefore, points $P$, $D$, $M$, and $N$ are coplanar, and $PD = 4MN$.