Let $P_1, P_2, \cdots, P_k$ be projection operators satisfying $\sum_{i=1}^{k} P_i = I$. The function $\Phi(X)$ is defined as follows: $$ \Phi(X) = \sum_{i=1}^{k} P_i X P_i$$ Show that $\Phi^2 = \Phi$.
My try: It can be shown that $P_iPj=0, i\neq j$, which gives: $$\Phi^2(X) = \sum_{i,j}P_i X P_iP_j X P_j = \sum_{i}P_i X P_i X P_i$$ I have no idea on how to proceed further.
Your computation of $\Phi^2(X)$ seems to be incorrect; informally, it shouldn't involve more than one $X$.
We should have
$$\Phi^2(X) = \Phi\left(\sum_{i = 1}^k P_i X P_i\right) = \sum_{j = 1}^k P_j \left(\sum_{i = 1}^k P_i X P_i\right) P_j .$$ Rearranging the sums and applying the definition of projection operator, together with the fact that you mentioned (that $P_i P_j = 0$ for $i \neq j$), yields the identity.