Answer
We can rank the situations according to the net electric potential at the origin:
$(b) \gt (a) = (c) = (d)$
Work Step by Step
We can write a general equation for the net electric potential:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
Let $r$ be the distance from each charge to the origin.
We can find the net electric potential at the origin for each situation:
(a) $V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{+2q}{r}-\frac{9q}{r}) = \frac{1}{4\pi~\epsilon_0}(-\frac{7q}{r})$
(b) $V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{-2q}{r}-\frac{3q}{r}) = \frac{1}{4\pi~\epsilon_0}(-\frac{5q}{r})$
(c) $V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{-2q}{r}-\frac{2q}{r}-\frac{2q}{r}-\frac{q}{r}) = \frac{1}{4\pi~\epsilon_0}(-\frac{7q}{r})$
(d) $V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{2q}{r}+\frac{2q}{r}-\frac{4q}{r}-\frac{7q}{r}) = \frac{1}{4\pi~\epsilon_0}(-\frac{7q}{r})$
We can rank the situations according to the net electric potential at the origin:
$(b) \gt (a) = (c) = (d)$
