Since $P Q R S$ is rectangular, then $\angle S R Q=\angle S P Q=90^{\circ}$. Also, $S R=P Q=20$ and $S P=Q R=15$. By the Pythagorean Theorem in $\triangle S P Q$, since $Q S>0$, we have $Q S=\sqrt{S P^{2}+P Q^{2}}=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625}=25$. Draw perpendiculars from $P$ and $R$ to $X$ and $Y$, respectively, on $S Q$. Also, join $R$ to $X$. We want to determine the length of $R P$. Now, since $\triangle S P Q$ is right-angled at $P$, then $\sin (\angle P S Q)=\frac{P Q}{S Q}=\frac{20}{25}=\frac{4}{5}$ and $\cos (\angle P S Q)=\frac{S P}{S Q}=\frac{15}{25}=\frac{3}{5}$. Therefore, $X P=P S \sin (\angle P S Q)=15\left(\frac{4}{5}\right)=12$ and $S X=P S \cos (\angle P S Q)=15\left(\frac{3}{5}\right)=9$. Since $\triangle Q R S$ is congruent to $\triangle S P Q$ (three equal side lengths), then $Q Y=S X=9$ and $Y R=X P=12$. Since $S Q=25$, then $X Y=S Q-S X-Q Y=25-9-9=7$. Consider $\triangle R Y X$, which is right-angled at $Y$. By the Pythagorean Theorem, $R X^{2}=Y R^{2}+X Y^{2}=12^{2}+7^{2}=193$. Next, consider $\triangle P X R$. Since $R X$ lies in the top face of the cube and $P X$ is perpendicular to this face, then $\triangle P X R$ is right-angled at $X$. By the Pythagorean Theorem, since $P R>0$, we have $P R=\sqrt{P X^{2}+R X^{2}}=\sqrt{12^{2}+193}=\sqrt{144+193}=\sqrt{337} \approx 18.36$. Of the given answers, this is closest to 18.4.