Explanation

(d)
Given, $\cos{\theta - 4\sin{\theta = 1}}$ …(i)
$\cos^{2}{\theta + 16\sin^{2}{\theta - 8\sin{\theta\cos{\theta = 1}}}}\ \ \$[on squaring]
$\Rightarrow \ \ 15\sin^{2}{\theta - 8\sin{\theta\ \cos{\theta = 0}}}$
$\Rightarrow \ \sin{\theta\left( 15\sin{\theta - 8\cos\theta} \right) = 0}$
$\Rightarrow \ \sin{\theta = 0}$ or $\tan{\theta = \frac{8}{15}}$
But $\tan\theta$ is not satisfy the Eq. (i),
$\therefore\ \sin{\theta = 0\ \ \Rightarrow \ \ \theta = 0,\ \pi}$
At $\theta = 0,\ \sin{\theta + 4\cos{\theta = 0 + 4 = 4}}$
At $\theta = \pi,\sin{\theta + 4\cos{\theta = 0 - 4 = - 4}}$