We can have $1$ bit or $2$ bit Booth codes. This question is about $2$ bit Booth codes. In Booth's algorithm, we get partial products by multiplying specific codes with the multiplicand. These partial products are used to get the actual product. We initially calculate $2$ bit booth code of the multiplier in this question. Then each bit of the code is multiplied with the multiplicand to get the partial product as shown in the last column of the given table.
Here, the multiplicand is $Y.$ So, notice that each row of partial product column is multiplied with $Y.$
Now, the question is how to get these codes i.e., how to represent a multiplier with a $2$ bit booth code. For that we need to look at the pair of $3$ bits as shown in the table below. To get code $C_{i},$ look for $3$ bits as shown.
$${\begin{array}{ccc|c}
\bf{x_{i+1}}& \bf{x_i}& \bf{x_{i+1}}&\bf{ Booth code (C_i)}\\\hline
0&0&0&0 \\ 0&0&1&1 \\ 0&1&0&1 \\ 0&1&1&2\\ 1&0&0&-2\\ 1&0&1&-1 \\ 1&1&0&- 1 \\ 1&1&1&0\\
\end{array}}$$
Now, multiply $i^{\text{th}}$ code to get partial product.
Therefore, Option C is correct.
