If the circles $$x^2 + y^2 + 2ax + cy + a = 0$$ and $$x^2 + y^2 - 3ax + dy - 1 = 0$$ intersect in two distinct points $$P$$ and $$Q$$ then the line $$5x + by - a = 0$$ passes through $$P$$ and $$Q$$ for

The two intersection points $$P$$ and $$Q$$ of the circles must lie on their common chord (radical axis).
For circles $$C_1: x^2 + y^2 + 2ax + cy + a = 0$$ and $$C_2: x^2 + y^2 - 3ax + dy - 1 = 0$$, subtract the second equation from the first:

$$\bigl(x^2 + y^2 + 2ax + cy + a\bigr) \;-\;\bigl(x^2 + y^2 - 3ax + dy - 1\bigr)=0$$
$$\Longrightarrow 5a\,x \;+\;(c-d)\,y \;+\;(a+1)=0$$

This straight-line equation is the radical axis, and hence the unique line through $$P$$ and $$Q$$.

The line given in the question is $$5x + by - a = 0$$. For this line to coincide with the radical axis, every corresponding coefficient must be proportional; that is, there must exist a non-zero constant $$k$$ such that

$$5 \;=\; k\,(5a), \qquad b \;=\; k\,(c-d), \qquad -a \;=\; k\,(a+1).\quad -(1) $$

From the first relation of $$-(1)$$, since $$5 \neq 0$$, we obtain $$k = \dfrac{1}{a}$$, which immediately forces $$a \neq 0$$.

Substitute $$k = \dfrac{1}{a}$$ into the third relation of $$-(1)$$:

$$-a \;=\;\dfrac{1}{a}\,(a+1) \;\;\Longrightarrow\;\; -a^2 = a + 1 \;\;\Longrightarrow\;\; a^2 + a + 1 = 0.$$

The quadratic $$a^2 + a + 1 = 0$$ has discriminant $$\Delta = 1 - 4 = -3 \lt 0$$, so it possesses no real roots.

Thus there exists no real value of $$a$$ for which the line $$5x + by - a = 0$$ can serve as the common chord of the two circles. Therefore, even though the circles intersect in two distinct points, that line never passes through both of them.

Option B which is: no value of $$a$$.