$(X*Y*Z =192$ and $Z=4)$ $\Rightarrow$ $X*Y= 48$
$\rightarrow$Given : $P = \frac{X+Y}{2}$
$\rightarrow$Average of $2$ numbers is minimum when the difference of that $2$ numbers is minimum.
$\rightarrow$We can write $48$ as product of $1*48,2*24,3*16,4*12,6*8$
$\rightarrow$Among all these $6*8$ is the expression which contains the numbers $6$ and $8$ and they have the minimum difference as compared to other expressions.
$\rightarrow$So $X=6$ and $Y=8$
$\Rightarrow$ $P = \frac{X+Y}{2} = \frac{6+8}{2} = 7 $
$\therefore$ Option B. $7$ is the correct answer.