Explanation

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