HMMT November, 2019

Solution:
Let $M$ be the foot of the altitude from $B$ to $CD$. Then $2CM + AB = CD \Longrightarrow CM = 3$. Then $DM = 4$ and by the Pythagorean theorem, $BM = 4$. Thus $BMD$ is a right isosceles triangle, i.e. $\angle BDM = \angle PDC = \frac{\pi}{4}$. Similarly, $\angle PCD = \frac{\pi}{4}$. Thus $\angle DPC = \frac{\pi}{2}$, which means quadrilateral $PQDC$ is cyclic. Now, $\sin \angle RPD = \sin \angle DCQ = \sin \angle MCB = \frac{4}{5}$.

Note that $AC \perp BD$ since $AB^{2} + CD^{2} = 1^{2} + 7^{2} = 5^{2} + 5^{2} = BC^{2} + DA^{2}$. Thus $P$ is the foot of the altitude from $D$ to $AC$. Since $D$ is on the circumcircle of $\triangle ABC$, line $PQR$ is the Simson line of $D$. Thus $R$ is the foot from $D$ to $AB$. Then from quadrilateral $RAPD$ being cyclic we have $\angle RPD = \angle RAD$. So $\sin \angle RPD = \frac{4}{5}$.