4. As shown in Figure 2, in the right trapezoid $M N P Q$, it is known that $\angle M N P=90^{\circ}, P M \perp N Q$. If $\frac{N Q}{P M}=\frac{\sqrt{2}}{2}$, then $\frac{M Q}{N P}$ $=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{1}{4}$
(D) $\frac{2}{3}$
4. A.

It is easy to prove $\triangle M Q N \backsim \triangle N M P$
$$
\begin{array}{l}
\Rightarrow \frac{M Q}{N M}=\frac{N Q}{P M}=\frac{M N}{N P} \\
\Rightarrow \frac{M Q}{N P}=\left(\frac{N Q}{P M}\right)^{2}=\frac{1}{2}
\end{array}
$$