Given NID=20 bits and HID=12 bits
hence, 131.61. 0000 | 0000.0000 0000 , the bits before the straight line ‘|’ shouldn’t be touched.
so in the remaining 12 bits we can subnet as required.
If we look carefully the required number is 1000.
Hence we can allocate a total of 1023 hosts by setting the 11th bit from the right end to 1 as we can set the remaining bits as required.
Hence the range of IP address would be 131.61.8.0/22 – 131.61.15.255/22.
Hence here we can select options A and C.