> Perform a bitwise AND with 0xFB and check if the result equals 0x09. Both 0x0D & 0xFB = 0x09 and 0x09 & 0xFB = 0x09.

This explanation was a bit unsatisfying. This works because 0x09 and 0x0D differ by a single bit, and 0xFB masks that bit (and only that bit) out.

If they differed by more than one bit, the fact that they & the same would be necessary but not sufficient.