A cool way to solve this problem is,
we have a 8 bits binary number which is in 2's complement form.
Now the 8th bit that is the MSB bit, represents the sign of the number.
So, 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1
now with 8 bits we have the total range as
10000000 = -128 (smallest number) to 01111111 = 127 (largest number)
So we have range as -128, -127, -126, -125,... 0, 1, 2, 3, 4, 5,... 127
as the last option is clearly the last largest number in our range therefore it's the answer.
Option D.