jmc

We have $\angle B=\angle PTQ$ and $\angle TPQ=180^{\circ }-\angle QPA-\angle APB=90^{\circ }-\angle APB=\angle BAP$. Therefore, $△BAP\sim △TPQ$. Since $AB/AP=4/5$, triangles $BAP$ and $PTQ$ are $\{3,4,5\}$ right triangles, and we have $TQ=\frac{3}{5}(15)=9$ and $TP=\frac{4}{5}(15)=12$. Since $ABCD$ is a rectangle and $TS$ is perpendicular to $BC$, then $ABTS$ is also a rectangle. Thus, $TS=BA=16$ and $QS=TS-QT=16-9=7$.

In triangles $PQT$ and $DQS$, $\angle PTQ=\angle DSQ=90^{\circ }$. Also, $\angle PQT$ and $\angle DQS$ are vertically opposite angles and are therefore equal. Therefore, $△PQT$ and $△DQS$ are similar triangles. Since $△PQT$ and $△DQS$ are similar triangles, the ratios of corresponding side lengths in these two triangles are equal.

That is, $\frac{SD}{TP}=\frac{QS}{QT}$ or $\frac{SD}{12}=\frac{7}{9}$ or $SD=12\times \frac{7}{9}=\boxed{\frac{28}{3}}$.