Given equation of the parabola is $y^2=4ax$
Differentiating w.r.to $x$, we get $2y\large\frac{dy}{dx}$$=4a$
$\Rightarrow\:\large\frac{dy}{dx}=\frac{2a}{y}$
Slope of tangent to this curve at $ (x_1,y_1)$ is $\large\frac{dy}{dx}=\frac{2a}{y_1}$
Slope of normal$=\large\frac{-1}{dy/dx}$$=-\frac{y_1}{2a}$
$\therefore$ For the parabola $y^2=4ax$ equation of normal at $(x_1,y_1)$ is
$(y-y_1)=-\large\frac{-y_1}{2a}$$(x-x_1)$
Parametric coordinates of the parabola is given by $(x_1,y_1)=(at^2,2at)$
$\therefore$ Eqn. of normal becomes $(y-2at)=-\large\frac{2at}{2a}$$(x-at^2)$
$\Rightarrow\:y-2at=-x+at^3$
$\Rightarrow\:y+x=2at+at^3$
Hence (a) is the correct answer.