Normals drawn to y 2 = 4 a x at the points where it is intersected by the line y = m x + c intersected at P …

Let $y=mx+c$, intersect $y^2=4ax$ at $A\left(a{t}_1^2,2a{t}_1\right)$ and $B\left(a{t}_2^2,2a{t}_2\right)$.

Then, $\frac{2}{t_1+t_2}=m$

$\Rightarrow t_1+t_2=\frac{2}{m}$

Let the foot of another normal be $C\left(a{t}_3^2,2a{t}_3\right)$

Then, $t_1+t_2+t_3=0$

$t_3=\left(t_1+t_2\right)=-\frac{2}{m}$

Thus, other foot is $\left(\frac{4a}{m^2},\frac{-4a}{m}\right)$

Hence, option (d) is correct.