Let A = [ cases m & n p & q cases ], d = |A| ≠ 0 and |A - d(Adj A)|...

Given, A = \left[\begin{cases}m & n \\ p & q\end{cases}\right], \left|A - d\left(adj A\right)\right| = 0

Here, $d = | A | = mq - np$

⇒ \left|A - d\left(adj A\right)\right| = \left|\left[\begin{cases}m & n \\ p & q\end{cases}\right] - d\left[\begin{cases}q & - n \\ - p & m\end{cases}\right]\right| = 0

⇒ \left|\begin{cases}m - qd & n\left(1 + d\right) \\ p\left(1 + d\right) & q - md\end{cases}\right| = 0

$⇒ \left(m - qd\right)\left(q - md\right) - np{\left(1 + d\right)}^{2} = 0$

$⇒ mq - {m}^{2}d - {q}^{2}d + mq{d}^{2} - np\left(1 + 2d + {d}^{2}\right) = 0$

$⇒ mq - {m}^{2}d - {q}^{2}d + mq{d}^{2} - np - 2npd - np{d}^{2} = 0$

$⇒ \left(mq - np\right) + {d}^{2}\left(mq - np\right) - d\left({m}^{2} + {q}^{2} + 2np\right) = 0$

$⇒ ( mq - np ) ( 1 + {d}^{2} ) - d ( m + q{)}^{2} = 0$

$⇒ 1 + {d}^{2} = \dfrac{d ( m + q{)}^{2}}{( mq - np )}$

$⇒ 1 + {d}^{2} = \dfrac{( mq - np ) ( m + q{)}^{2}}{( mq - np )}$\left[∵ d = mq - np\right]$</p><p>$⇒ {\left(1 + d\right)}^{2} = {\left(m + q\right)}^{2}$</p><p>Hence, option \left(A\right)$ is correct.