2. As shown in Figure 1, in rectangle $A B C D$, $A B=1, B C$ $=m, O$ is the center of the rectangle, $P O \perp$ plane $A B C D, P O=n$, and there exists a unique point $E$ on side $B C$ such that $P E \perp D E$. If the angle between plane $P D E$ and plane $A B C D$ is $60^{\circ}$, then $(m, n)$ is ( ).
(A) $\left(2, \frac{3}{2}\right)$
(B) $\left(2 \sqrt{2}, \frac{3}{2}\right)$
(C) $\left(\frac{3}{2}, 2\right)$
(D) $\left(\frac{3}{2}, 2 \sqrt{2}\right)$
2. B.

As shown in Figure 7, let $C E = x$, and draw $O F \perp B C$ at $F$. Then
$$
\begin{array}{l}
O F = \frac{1}{2}, \\
E F = \frac{m}{2} - x .
\end{array}
$$

Thus, $P E^{2} = P O^{2} + O F^{2} + E F^{2}$
$$
\begin{array}{l}
= n^{2} + \frac{1}{4} + \left(\frac{m}{2} - x\right)^{2} \\
= x^{2} - m x + \frac{m^{2}}{4} + n^{2} + \frac{1}{4}, \\
D E^{2} = C E^{2} + C D^{2} = x^{2} + 1, \\
P D^{2} = P O^{2} + O D^{2} \\
= n^{2} + \left(\frac{\sqrt{1 + m^{2}}}{2}\right)^{2} = \frac{1}{4} m^{2} + n^{2} + \frac{1}{4} .
\end{array}
$$

Since $P E \perp D E$, we have
$$
P E^{2} + D E^{2} = P D^{2},
$$

which simplifies to $2 x^{2} - m x + 1 = 0$.
Given that point $E$ exists and is unique, then $\Delta = m^{2} - 8 = 0$, yielding $m = 2 \sqrt{2}$. Thus, $x = \frac{\sqrt{2}}{2}$.
Since $P O \perp$ plane $A B C D$ and $P E \perp D E$, it follows that $O E \perp D E$.
Therefore, $\angle P E O$ is the angle between plane $P D E$ and plane $A B C D$, i.e., $\angle P E O = 60^{\circ}$.
In the right triangle $\triangle P O E$, we have
$$
\sin \angle P E O = \frac{P O}{P E} = \frac{\sqrt{3}}{2},
$$

which gives $n = \frac{3}{2}$.