Explanation

(d)
The equation of any normal to $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is
$ax\sec{\phi - by\ cosec\ \phi = a^{2} - b^{2}\ \ \ \ \ \ \ \ \ldots(i)}$
Given straight line $x\cos{\alpha + y\sin\alpha = p}$ will be a normal to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ if Eq.(i)and $x\cos\alpha + y\sin\alpha = p$ represent the same line.
$\frac{{a\ sec}\phi}{\cos\alpha} = - \frac{b\ cosec\ \phi}{\sin\alpha} = \frac{a^{2} - b^{2}}{p}$
$\Rightarrow \ \cos\phi = \frac{ap}{\left( a^{2} - b^{2} \right)\cos\alpha}\ ,$
$\sin\phi = \frac{- bp}{\left( a^{2} - b^{2} \right)\sin\alpha}$
$\therefore\ \sin^{2}{\phi + \cos^{2}{\phi = 1}}$
$\Rightarrow \ \ \frac{b^{2}p^{2}}{\left( a^{2} - b^{2} \right)^{2}\sin^{2}{\alpha\ }} + \frac{a^{2}p^{2}}{\left( a^{2} - b^{2} \right)^{2}\cos^{2}{\alpha\ }} = 1$
$\Rightarrow \ \ p^{2}(b^{2}cosec^{2}\alpha + a^{2}\sec^{2}{\alpha) = \left( a^{2} - b^{2} \right)^{2}}$