Let Variance of X = p
$\sigma _{x}$ = $\sqrt{p}$
Variance of Y = $a^{2}p$
($\because$ When we add a constant to each element of the list, the standard deviation (or variance) remains unchanged. This is because, the mean also gets added by the same constant and hence the deviation from the mean remains the same for each element. when we multiply a random variable with a constant, since we square the values while calculating variance, the value of a(constant) will also gets squared)
$\sigma _{y}$ = $a\sqrt{p}$ = a * $\sigma _{x}$
$\therefore $ $\sigma _{y}$ = $a \sigma _{x}$
Answer D is incorrect