quantum mechanics - Why is the projection operator corresponding to $tilde M$ given by $P_motimes I_B$?

Nielsen and Chuang, Chapter 2 (Box 2.6):

Suppose $M$ is any observable on a system $A$, and we have some measuring device which is capable of realizing measurements of $M$. Let $\tilde M$ denote the corresponding observable for the same measurement, performed on the composite system $AB$. Our immediate goal is to argue that $\tilde M$ is necessarily equal to $M \otimes I_B$. Note that if the system $AB$ is prepared in the state $|m\rangle |\psi\rangle$, where $|m\rangle$ is an eigenstate with an eigenvalue $m$ and $|\psi\rangle$ is any state of $B$, then the measuring device must yield the result $m$ for the measurement, with probability one. Thus, if $P_m$ is the projector onto the $m$ eigenspace of the observable $M$, then the corresponding projector for $\tilde M$ is $P_m \otimes I_B$. We therefore have

$\tilde M = \sum_{m} m P_m\otimes I_B = M \otimes I_B$

Could someone please explain me why the projection operator is $ P_m\otimes I_B$? A proof or at least an example which illustrates the motivation behind the formula would be helpful. It has not been explained clearly in the textbook.