Explanation

(d)
Since, $a + b = - p,\ ab = 1$ ...(i)
And $c + d = - q,\ cd = 1$
Now, $(a - c)(b - c)$ and $(a + d)(b + d)$ are the roots of $x^{2} + ax + \beta = 0$
$\therefore\ \ \ (a - c)(b - c)(a + d)(b + d) = \beta$
$\Rightarrow \ \ \left( ab - ac - bc + c^{2} \right)\left( ab + ad + bd + d^{2} \right) = \beta$
$\Rightarrow \ \ \left\{ 1 - c(a + b) + c^{2} \right\}\left\{ 1 + d(a + b) + d^{2} \right\} = \beta$
$\Rightarrow \ \ \ \ \left( 1 + pc + c^{2} \right)\left( 1 - pd + d^{2} \right) = \beta$
$\Rightarrow \ \ 1 - pd + d^{2} + pc - p^{2}cd + pcd^{2} + c^{2} - pc^{2}d + c^{2}d^{2} = \beta$
$\Rightarrow \ \ 1 - pd + d^{2} + pc - p^{2} + pd + c^{2} - pc + 1 = \beta$
$\lbrack\because\ \ cd = 1\rbrack$
$\Rightarrow \ \ 2 + d^{2} + c^{2} - p^{2} = \beta$
$\Rightarrow \ \ 2cd + c^{2} + d^{2} - p^{2} = \beta$ [$\because 1 = cd$]
$\Rightarrow \ \ (c + d)^{2} - p^{2} = \beta$
$\Rightarrow q^{2} - p^{2} = \beta$ ($\because\ \ c + d = - q$)