If the positive number d is the standard deviation of n, k and p then the standard deviation of n+1, k+1, and p+1 is...

The standard deviation of a set of numbers will not change if the same number is added to every number of the set. Assuming that n, k, p = 1, 2, 3 respectively
$$SD=\sqrt{\frac{\left(1-2\right)^2+\left(2-2\right)^2+\left(3-2\right)^2}{3}}=\sqrt{\frac{1+0+1}{3}}=\sqrt{\frac{2}{3}}$$
$$for\ n+1,\ k+1,\ and\ p+1\ =>\ 2,\ 3,\ 4$$
$$SD=\sqrt{\frac{\left(2-3\right)^2+\left(3-3\right)^2+\left(4-3\right)^2}{3}}=\sqrt{\frac{1+0+1}{3}}=\sqrt{\frac{2}{3}}$$
So the standard deviation d for n, p, and k remains unchanged for n+1, p+1, and k+1

$$Answer\ =\ E$$