$D$ is a point on side $QR$ of $\triangle PQR$ such that $PD \perp QR$. Will it be correct to say that $\triangle PQD \sim \triangle RPD$? Why?

(N/A) No,it is not correct to say that $\triangle PQD \sim \triangle RPD$ in general.
In $\triangle PQD$ and $\triangle RPD$:
$1$. $\angle PDQ = \angle PDR = 90^{\circ}$ (Given that $PD \perp QR$)
$2$. $PD = PD$ (Common side)
For two triangles to be similar,we need either $AA$,$SAS$,or $SSS$ similarity criteria. Here,we only have one angle and one side equal. We do not have information about the equality of other angles or the proportionality of other sides.
Therefore,$\triangle PQD$ is not necessarily similar to $\triangle RPD$ unless $\triangle PQR$ is a specific type of triangle (e.g.,if $\angle P = 90^{\circ}$ and $PD$ is the altitude to the hypotenuse,then $\triangle PQD \sim \triangle RPD$ by $AA$ similarity).