Computing $H(Z|B)$ in a bipartite density matrix $rho_{AB}$

Let’s say Bob prepares a bipartite quantum state $rho_{AB}$ to be shared between him and Alice. Bob sends Alice’s part to her lab. Alice measures her subsystem $A$ in the computational basis $mathcal{Z}$. Now we want to know the uncertainty in Alice’s system given Bob’s quantum memory register($B$). This quantity is denoted as $H(Z|B)$.

I was reading this paper by Berta et al. where they mention in a footnote (page 2, footnote 4) that $H(R|B)$ is the conditional von Neumann entropy of the following state:
$$
left(sum_j |psi_j rangle langlepsi_j| otimes mathbb{1} right) rho_{AB} left(sum_j |psi_j rangle langlepsi_j| otimes mathbb{1} right),
$$
where $|psi_j rangle$ is the eigenvector of the measurement $mathcal{R}$. In our case, this would be $mathcal{Z}$.

My confusion now is this. We know that the sum of eigenvectors of the computational basis $mathcal{Z}$ makes up the identity matrix $mathbb{1}$ again. In this case, isn’t it becoming the following?

$$
left(mathbb{1} otimes mathbb{1} right) rho_{AB} left( mathbb{1} otimes mathbb{1} right) = rho_{AB},
$$
in which case the conditional von Neumann entropy is simply $H(A|B)$. In the end, are we getting $H(Z|B) = H(A|B)$? This does not seem correct though. Thanks in advance.