$P$ and $Q$ are isogonal conjugates w.r.t. $\triangle ABC\implies$ midpoint of $PQ$ is circumcentre of $\triangle DEF$. But $\angle DEF=90^{\circ}\implies$ circumcentre of $\triangle DEF$ is midpoint of $DE$. $\therefore$ $PQ$ and $DE$ bisect each other $\implies PDQE$ is a parallelogram. $\therefore$ $PD\perp DB\implies EQ\perp BD$ and similarly $DQ\perp BE$. $\therefore Q$ is orthocentre of $\triangle BDE$ as desired. Q.E.D.

how does P and Q being isogonal conjugates wrt to ABC imply that the midpoint of PQ is the circumcenter of DEF?

It is well known that the midpoint of isogonal conjugates w.r.t. to a triangle is the circumcentre of their pedal triangles w.r.t. to the same triangle.

Angle chasing gives $\angle AQC = 90$. This gives $\triangle AFQ$, $\triangle APC$, and $\triangle QDC$ similar by spiral similarity. And so, from here we know that $FQ$ = $PD$ and $FP = QD$. So, $PDQF$ is a parallelogram as desired.