(P) ${\sin }^{6}x+{\cos }^{4}x=1$. Since ${\sin }^{6}x\le {\sin }^{2}x$ and ${\cos }^{4}x\le {\cos }^{2}x$ with sum ${\sin }^{2}x+{\cos }^{2}x=1$, equality holds only when both inequalities are equalities. This forces ${\sin }^{2}x\in \{0,1\}$ and ${\cos }^{2}x\in \{0,1\}$ simultaneously, giving $\sin x=0$ or $\cos x=0$. In $[-\pi ,\pi ]$: $x\in \{-\pi ,-\pi /2,0,\pi /2,\pi \}$. 5 elements. P $\to$ 5.

(Q) ${\sin }^{2}x+{\cos }^{6}x=1$. The same reasoning gives ${\cos }^{2}x\in \{0,1\}$. In $[-\pi /2,\pi /2]$: $x\in \{-\pi /2,0,\pi /2\}$. 3 elements. Q $\to$ 3.

(R) ${\cos }^2(x/2)-{\sin }^{2}x=\frac{1}{2}$ rewrites as $\frac{1+\cos x}{2}-(1-{\cos }^{2}x)=\frac{1}{2}$, giving $2{\cos }^{2}x+\cos x-2=0$.

$\cos x=\frac{-1\pm \sqrt{17}}{4}$. Only $\frac{-1+\sqrt{17}}{4}\approx 0.78$ lies in $[-1,1]$; the other root $\approx -1.28$ is outside. So $\cos x=\frac{-1+\sqrt{17}}{4}$ has exactly two solutions in $[-\pi ,\pi ]$ (symmetric pair). 2 elements. R $\to$ 2.

(S) $6{\sin }^2(x/2)-\cos 3x=3$. Using $2{\sin }^2(x/2)=1-\cos x$, the equation becomes $3(1-\cos x)-\cos 3x=3$, i.e. $\cos 3x+3\cos x=0$. Using $\cos 3x=4{\cos }^{3}x-3\cos x$: $4{\cos }^{3}x=0\Rightarrow \cos x=0$. In $[-2\pi ,2\pi ]$: $x\in \{-3\pi /2,-\pi /2,\pi /2,3\pi /2\}$. 4 elements. S $\to$ 4.

$\therefore$ The correct option is (B).