The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is

The Henderson-Hasselbalch equation for a weak acid buffer is
$$\text{pH} = \text{p}K_a + \log\frac{[\!A^-]}{[\!HA]}$$

If 50 % of the acid $$HA$$ is ionised, then starting from any initial concentration $$C$$:
$$[\!A^-] = 0.5\,C \quad\text{and}\quad [\!HA] = 0.5\,C$$

Hence the ratio
$$\frac{[\!A^-]}{[\!HA]} = \frac{0.5\,C}{0.5\,C} = 1$$ and $$\log 1 = 0$$.

Given $$\text{p}K_a = 4.5$$, we get
$$\text{pH} = 4.5 + 0 = 4.5$$.

The relation between pH and pOH is $$\text{pH} + \text{pOH} = 14$$, therefore
$$\text{pOH} = 14 - 4.5 = 9.5$$.

Option C which is: 9.5