$\lim\limits_{x\to 0}\large\frac{\sin(\pi\cos^2x)}{x^2}=$$\lim\limits_{x\to 0}\large\frac{\sin(\pi-\pi\sin^2x)}{x^2}$
$[\sin(\pi-\theta)=\sin\theta]$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\sin(\pi\sin^2x)}{\pi\sin^2x}\times \frac{\pi\sin^2x}{x^2}$
$\Rightarrow \pi$
Hence (b) is the correct answer.