Multiplication can be directly carried in $2$’s complement form. $\textsf{F87B} = 1111 1000 0111 1011$ can be left shifted $3$ times to give $8P = 1100 0011 1101 1000 = \textsf{C3D8}.$
Or, we can do as follows:
MSB in $\textsf{(F87B)}$ is $1.$ So, $P$ is a negative number. So, $P = -1 \ast 2$'s complement of $\textsf{(F87B)} = -1 \ast (0785) = -1 \ast (0000 0111 1000 0101)$
$8 \ast P = -1 \ast (0011 1100 0010 1000) (P$ in binary left shifted $3$ times$)$
In $2$'s complement representation , this equals, $1100 0011 1101 1000 = \textsf{C3D8}$
Correct Answer: $A$