The stack in the 8085 is a region of RAM used for temporary storage of register-pair contents and return addresses, and it grows downward (toward lower addresses). The Stack Pointer (SP) is a 16-bit register that always points to the top of the stack.

The PUSH rp instruction saves a 16-bit register pair (such as B-C, D-E, H-L, or PSW) onto the stack. Because each register pair is two bytes, the operation proceeds in two steps:

1. SP is decremented by 1, and the high-order byte of the register pair is stored at the address now pointed to by SP.
2. SP is decremented by 1 again, and the low-order byte is stored at the new SP location.

After both steps, the net change is that SP has been decremented by 2, which is the correct answer. This reflects the fact that two bytes were pushed and the stack top moved down by two addresses.

Why the other options are wrong:

• Incremented by 2 is exactly what happens during a POP (which removes two bytes and moves the pointer back up), the inverse of PUSH.
• Incremented by 1 would correspond to storing only a single byte upward, which does not match a two-byte push on a downward-growing stack.
• Unchanged is impossible, since PUSH must move the pointer to make room for the two bytes it stores.