The lines $$p(p^2 + 1)x - y + q = 0$$ and $$(p^2 + 1)^2 x + (p^2 + 1)y + 2q = 0$$ are perpendicular to a common line for

The general form of a straight line is $$Ax + By + C = 0$$, whose slope is $$m = -\dfrac{A}{B}$$ (provided $$B \neq 0$$).

For the first line
$$p\bigl(p^{2}+1\bigr)x - y + q = 0$$
we have $$A_{1} = p(p^{2}+1)$$ and $$B_{1} = -1$$, so its slope is
$$m_{1} = -\dfrac{A_{1}}{B_{1}} = -\dfrac{p(p^{2}+1)}{-1} = p(p^{2}+1).$$

For the second line
$$(p^{2}+1)^{2}x + (p^{2}+1)\,y + 2q = 0$$
we have $$A_{2} = (p^{2}+1)^{2}$$ and $$B_{2} = (p^{2}+1)$$, so its slope is
$$m_{2} = -\dfrac{A_{2}}{B_{2}} = -\dfrac{(p^{2}+1)^{2}}{(p^{2}+1)} = -(p^{2}+1).$$

Let a third line with slope $$m$$ be perpendicular to both given lines. For perpendicularity we need

$$m\,m_{1} = -1 \quad\text{and}\quad m\,m_{2} = -1.$$ Cancelling $$m$$ gives $$m_{1} = m_{2}.$$

Therefore the two given lines themselves must be parallel:

$$p(p^{2}+1) = -(p^{2}+1).$$

Factorising,

$$(p^{2}+1)(p+1) = 0.$$

Since $$p^{2}+1 \gt 0$$ for all real $$p$$, the only real solution is

$$p = -1.$$

For $$p = -1$$ both lines have finite, equal slopes, so a unique line with slope $$m = -\dfrac{1}{m_{1}}$$ is perpendicular to them.

Hence there is exactly one real value of $$p$$ for which the two given lines are perpendicular to a common line.

Option B which is: exactly one value of $$p$$