1. We need to evaluate

L=\lim_{x\to 0}\frac{\sin^2\!ig(\pi \cos^4 x\big)}{x^4}.

2. As $x\to 0$, we know

$$\cos x = 1-\frac{x^2}{2}+o(x^2).$$

So,

$$\cos^2 x = 1-x^2+o(x^2),$$

and hence

$$\cos^4 x = (\cos^2 x)^2 = (1-x^2+o(x^2))^2 = 1-2x^2+o(x^2).$$

A more standard expansion gives

$$\cos^4 x = 1-2x^2+O(x^4).$$

Therefore,

$$\pi\cos^4 x = \pi-2\pi x^2+O(x^4).$$

3. Now use

$$\sin(\pi-h)=\sin h.$$

Let

$$h=2\pi x^2+O(x^4).$$

Then

$$\sin(\pi\cos^4 x)=\sin\big(\pi-h\big)=\sin h.$$

For small $h$,

$$\sin h \sim h.$$

So,

$$\sin(\pi\cos^4 x) \sim 2\pi x^2.$$

4. Squaring,

$$\sin^2(\pi\cos^4 x) \sim (2\pi x^2)^2 = 4\pi^2 x^4.$$

Hence,

$$L=\lim_{x\to 0}\frac{\sin^2(\pi\cos^4 x)}{x^4}=4\pi^2.$$

5. Therefore the correct option is

$$\boxed{C:~4\pi^2}.$$