Since $PD$ and $CB$ are both perpendicular to $\ell_A$, we have that $PD \parallel CB$. Similarly, $PE \parallel CA$ and $PF \parallel BA$. Since $\angle PDH = \angle PEH = 90^\circ$, we have that $PDEH$ is cyclic, and $PH$ is the diameter of this circle. Similarly, $\angle PDH = \angle PFH = 90^\circ$, and so $DPFH$ is cyclic with $PH$ as diameter. It follows that $DPFHE$ is cyclic with $PH$ as diameter. We then have that $\angle FDE = \angle FPE = \angle BAC$ since $PE \parallel CA$ and $PF \parallel BA$. Similarly, we have that $\angle EFD = \angle ACB$, and so we have that triangle $ABC$ is similar to triangle $DEF$. Since the circumradius of $ABC$ is $R$, and the circumradius of triangle $DEF$ is $\frac{1}{2} PH$, the ratio of similarity is $\frac{PH}{2R}$. It follows that $$ \operatorname{area}(DEF) = \frac{PH^2}{4R^2} \cdot \operatorname{area}(ABC). $$