We have $4$ blocks and $2$ blocks in a set
$\implies$ there are $2$ sets. So blocks will go to sets as follows:
$$\begin{array}{|c|c|} \hline \textbf {Set Number} & \textbf{Block Number} \\\hline \text{0} & \text{0,8,12} \\\hline\text{1} & \\\hline \end{array}$$
Since the lowest bit of block address is used for indexing into the set, so $8, 12$ and $0$ first miss in cache with $0$ replacing $8$ (there are two slots in each set due to $2-\text{way}$ set) and then $12$ hits in cache and $8$ again misses. So, totally $4$ misses.
Correct Answer: $C$