Let the strategy of player 1 be represented by $(x1_1,xDD_1,xDC_1,xCD_1,xCC_1)$ where $x1$ is the first round action of player 1, $xDD_1$ is the action taken at the information set where both players have defected in the first round, $xDC_1$ is the action taken at the information set where player 1 has defected and player 2 has cooperated in round 1, etc.. Note that something like $(x1_1,x2_1)$ (with $x2_1$ being the action taken in round 2) is never a full specification of the strategy of player 1, since we need to specify behavior at each information set separately. Define the strategies of player 2 similarly. However, a perfect Bayesian equilibrium must also specify the beliefs of the player, $\mu_1,\mu_2$. This is an important part of the specification of an equilibrium. As we will see below, the question is geared towards understanding that a different equilibrium does not require the strategies to differ. A difference in beliefs is sufficient to count as a different equilibrium.
The perfect equilibrium is given by: $((D,D,D,D,D),\mu_1)$ for player 1 and $((D,D,D,D,D),\mu_2)$ for player 2, where $\mu_1$ and $\mu_2$ are consistent beliefs at all information sets.
As has been noted in the comments, since "defect" is a dominated strategy irrespective of beliefs, even in a weak perfect Bayesian equilibrium the strategy profiles must be $(D,D,D,D,D)$ for both players. However, the following is now also a weak perfect Bayesian Nash equilibrium: $((D,D,D,D,D),\mu_1')$ and $((D,D,D,D,D),\mu_2')$ with $\mu_1'$, $\mu_2'$ consistent on the equilibrium path.
Thus, the question is not wrong, it simply shows that two weak perfect Bayesian Nash equilibria can have identical strategies as long as they differ in beliefs off the equilibrium path.