$f\left(x\right)=\sqrt{3}\sin x-\cos x+2=2\sin \left(x-\frac{\pi }{6}\right)+2$

Since $f\left(x\right)$  is one-one and onto, $f$  is invertible.

Now $fo{f}^{-1}\left(x\right)=x$

$\Rightarrow$  $2\sin \left(f^{-1}\left(x\right)-\frac{\pi }{6}\right)+2=x$

$\Rightarrow \sin \left(f^{-1}\left(x\right)-\frac{\pi }{6}\right)=\frac{x}{2}-1$

$\Rightarrow f^{-1}\left(x\right)={\sin }^{-1}\left(\frac{x}{2}-1\right)+\frac{\pi }{6}$

Because $\left|\frac{x}{2}-1\right|\le 1$  for all $x\in \left[0,4\right]$ .

Hence (B) is a correct answer.

Also using ${\sin }^{-1}\alpha +{\cos }^{-1}\alpha =\pi /2$

$f^{-1}\left(x\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{x-2}{2}\right)-\frac{\pi }{6}=\frac{2\pi }{3}-{\cos }^{-1}\left(\frac{x-2}{2}\right)$ .

Thus (C) is also correct.