Let $P : E \to E$ be a projection. $\DeclareMathOperator{\Tr}{Tr}$$\DeclareMathOperator{\Ima}{Im}$$\DeclareMathOperator{\rank}{rank}$ Notice that $P(P-1) = P^2 - P = 0$ so the only eigenvalues of $P$ can be $0$ and $1$. Furthermore, if $P\ne 0, I$ then $\mu_P(x) = x(x-1)$ is precisely the minimal polynomial of $P$, so $P$ is diagonalizable since $\mu_P$ consists only of linear factors.
This implies that $\Tr P = \rank(P)$, since both will count the number of ones on the diagonal of the diagonalized $P$.
Back to the problem, from $P_1 + \ldots + P_k = I$ we get that
$$E = \sum_{i=1}^k \Ima P_i$$
It suffices to prove that the sum is direct:
$$\dim E = \Tr I = \Tr (P_1 + \ldots + P_n) = \sum_{i=1}^k \Tr P_i = \sum_{i=1}^k \rank(P_i) = \sum_{i=1}^k \dim(\Ima P_i)$$
Now assume $\Ima P_{j_1} \cap \Ima P_{j_2} \ne \{0\}$ for some $j_1 \ne j_2$:
\begin{align}\dim E &= \dim\left(\sum_{i=1}^k \Ima P_i\right) \\
&\le \sum_{i\ne j_1,j_2} \dim\Ima P_i + \dim(\Ima P_{j_1} + \Ima P_{j_2}) \\
&= \sum_{i\ne j_1,j_2} \dim\Ima P_i + \dim\Ima P_{j_1} + \dim\Ima P_{j_2} - \underbrace{\dim(\Ima P_{j_1} \cap \Ima P_{j_2})}_{>0} \\
&< \sum_{i=1}^k \dim\Ima P_i\\
&= \dim E
\end{align}
This is a contradiction. Thus, the sum is indeed direct.
Now, take $i \ne j$. We have
$$P_i(\underbrace{P_jx}_{\in\Ima P_j}) = 0$$
since $\Ima P_i \cap \Ima P_j = \{0\}$. Therefore, $P_iP_j = 0$.
Note: This question was very helpful.
