Option $(A)$ RRC a, #1. As the 8 bit register is rotated via carry 8 times.
$a_7a_6a_5a_4a_3a_2a_1a_0$
$c_0a_7a_6a_5a_4a_3a_2a_1$, now $a_0$ is the new carry. So, after next rotation,
$a_0c_0a_7a_6a_5a_4a_3a_2$
So, after 8 rotations,
$a_6a_5a_4a_3a_2a_1a_0c_0$ and carry is $a_7$.
Now, one more rotation will restore the original value of $A_0$.