Whenever, $b_{2}$ = $b_{3}$$=$ $1$, then only $ 0100$ i.e., $4$ is added to the given binary number. Let's write all possibilities for $b$.
$${\begin{array}{|c|c|c|c|}\hline
\bf{b_3}& \bf{b_2}& \bf{b_1}&\bf{ b_0} \\\hline
0&0&0&0 \\ 0&0&0&1\\ 0&0&1&0 \\ 0&0&1&1 \\ 0&1&0&0 \\ 0&1&0&1 \\ 0& 1&1&0 \\ 0&1&1&1\\ 1&0&0&0\\ 1&0&0&1\\ 1&0&1&0 \\ 1&0&1&1 \\ 1&1&0&0 \\ 1& 1&0&1 \\ 1&1&1&0\\ 1&1&1&1 \\\hline
\end{array}}$$
Note that the last $4$ combinations ($1100$, $1101$, $1110$, $1111$) leads to $b_{3}$ and $b_2$ as $1$. So, in these combinations only $0100$ will be added.
$1100$ is $12$
$1101$ is $13$
$1110$ is $14$
$1111$ is $15$
in binary unsigned number system.
$1100 + 0100 = 10000$.
$1101 + 0100 = 10001$ and so on.
This is conversion to radix $12$.
Correct Answer: $D$