At first we have to translate the given virtual addresses (which addresses a byte) to page addresses (which again is virtual but addresses a page). This can be done simply by dividing the virtual addresses by page size and taking the floor value (equivalently by removing the page offset bits). Here, page size is $16$ bytes which requires $4$ offset bits. So,
$0, 4, 8, 20, 24, 36, 44, 12, 68, 72, 80, 84, 28, 32, 88, 92 \implies \ 0,0,0,1,1,2,2,0,4,4,5,5,1,2,5,5$
We have $4$ spaces for a page and there will be a replacement only when a $5^{th}$ distinct page comes. Lets see what happens for the sequence of memory accesses:$$\small \begin{array}{|c|c|c|clc|}\hline\textbf{Incoming}&\textbf{Page Address} & \textbf{No. of Page Faults}& \textbf{Pages in Memory}\\
\textbf{Virtual Address}&&&\textbf{in LRU Order} \\ \hline
0 & 0&1&0 \\ \hline
4 & 0&1&0 \\ \hline
8 & 0&1&0 \\ \hline
20 & 1&2&0,1 \\ \hline
24 & 1&2&0,1 \\ \hline
36 & 2&3&0,1,2 \\ \hline
44 & 2&3&0,1,2 \\ \hline
12 & 0&3&1,2,0 \\ \hline
68 & 4&4&1,2,0,4 \\ \hline
72 & 4&4&1,2,0,4 \\ \hline
80 & 5&5&2,0,4,5 \\ \hline
84 & 5&5&2,0,4,5 \\ \hline
28 & 1&6&0,4,5,1 \\ \hline
32 & 2&7&4,5,1,2 \\ \hline
88 & 5&7&4,1,2,5 \\ \hline
92 & 5&7&4,1,2,5 \\ \hline
\end{array}$$So, (B) choice.