Solution:

Let us denote by $E$ and $F$, respectively, the feet of the perpendiculars from $P$ and $Q$ to line $B C$, and by $M$ the midpoint of segment $P Q$.

Consider the point $X$ on side $B C$ such that $\text{B A X=2 B A P}$. Then also $\text{C A X= B A C-2 B A P=2 C A Q}$, so $P$ and $Q$ are, respectively, the incenters of triangles $B A X$ and $C A X$. It follows that $X P$ and $X Q$ are the bisectors of angles $B X A$ and $C X A$, so $\text{P X Q=90}$.

The condition $\text{P D Q=90}$ is equivalent to $M D=M P=M Q=M X$, and since $M E=M F$, it suffices to prove that $D E=X F$. Both lengths are easily computed on the basis of the "big problem":

$D E=B D-B E=\frac{A B+B C-A C}{2}-\frac{A B+B X-A X}{2}=\frac{C X-A C+A X}{2}=X F$.

Second solution. Let line $B I$ intersect the circumcircle of $\triangle A P Q$ again at point $N$. We have $\text{A I N=180 - B I A=90 - 2 = D I Q}$. Also, since $\text{I N Q= P A Q= 2}$ and $\text{I Q N= B I C- I N Q= (90 + 2 )- 2 =90}$, we have $\frac{I Q}{I N}=\sin \frac{\alpha}{2}=\frac{I A}{I D}$, whence $\frac{I A}{I N}=\frac{I D}{I Q}$. It follows that triangles $D I Q$ and $A I N$ are similar, so

$\text{I D Q= I A N=180 - A I N- A N I=180 - (90 - 2 )- A N P=90 + 2 - A Q P}$.

Analogously, $\text{I D P=90 + 2 - A P Q}$, so by adding we obtain

$\text{P D Q=180 + + 2 - (180 - P A Q )=90}$.