Detection of sulphide in solution also containing sulphate
I will add one thing that may be the reason behind why (C) is an acceptable answer choice. Yes, both $\ce{PbS}$ and $\ce{PbSO_4}$ are insoluble. However, if you look at the $\ce{K_{sp}}$ for both compounds, $\ce{PbS}$ is about $\pu{3.2 x10^-28}$ while the $\ce{K_{sp}}$ for $\ce{PbSO_4}$ is $\pu{1.3 x 10^-8}$. When you have a mixture of ions, the resulting molecule with the lowest $\ce{K_{sp}}$ will precipitate first. In this case, since $\ce{PbS}$ has a lower  $\ce{K_{sp}}$ than $\ce{PbSO_4}$, it will precipitate first.
This technique is called fractional precipitation where you separate two ions, in our case $\ce{S^{2-}}$ and $\ce{SO_4^{2-}}$, by precipitating one of the ions selectively through taking advantage of their differing $\ce{K_{sp}}$ values.
This is the likely reason behind why choosing (A) itself is acceptable or choosing (A) and (C) is also acceptable.