For matrix to be nonsingular any of these cases can occur$:$
$X_{1}=1,X_{2}=1,X_{3}=-1,X_{4}=1$ $ X1=-1,X2=1,X3=1,X4=1$
$X_{1}=1,X_{2}=-1,X_{3}=1,X_{4}=1$ $ X_{1}=-1,X_{2}=-1,X_{3}=-1,X_{4}=1$
$X_{1}=-1,X_{2}=1,X_{3}=-1,X_{4}=-1$ $ X_{1}=1,X_{2}=-1,X_{3}=-1,X_{4}=-1$
$X_{1}=-1,X_{2}=-1,X_{3}=1,X_{4}=-1 $ $X_{1}=1,X_{2}=-1,X_{3}=-1,X_{4}=-1$
each case has probability $=(\frac{1}{2})^{4} = \frac{1}{16}$
so required probability $=8\times\frac{1}{16}$ = $\frac{1}{2}$
So option $(A)$ is correct