Here is a slight variant of Jonas’ argument.
Assume that $ p_{1},\ldots,p_{n} $ are projection elements of a unital $ C^{*} $-algebra $ A $, where $ n \in \Bbb{N}_{\geq 2} $, such that
$$
\sum_{k = 1}^{n} p_{k} = 1_{A}.
$$
Choose distinct $ i,j \in [n] $, where $ [n] \stackrel{\text{df}}{=} \Bbb{N}_{\leq n} $. Then
\begin{align}
p_{i}
& = p_{i} 1_{A} \\
& = p_{i} \sum_{k \in [n]} p_{k} \\
& = \sum_{k \in [n]} p_{i} p_{k} \\
& = p_{i}^{2} + p_{i} p_{j} + \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} \\
& = p_{i} + p_{i} p_{j} + \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k}.
\end{align}
It follows that
$$
p_{i} p_{j} p_{j}^{*}
= p_{i} p_{j}
= - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k}
= - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} p_{k}^{*},
$$
and consequently,
$$
(\spadesuit) \qquad
(p_{i} p_{j}) (p_{i} p_{j})^{*}
= p_{i} p_{j} p_{j}^{*} p_{i}^{*}
= - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} p_{k}^{*} p_{i}^{*}
= - \sum_{k \in [n] \setminus \{ i,j \}} (p_{i} p_{k}) (p_{i} p_{k})^{*}.
$$
On the extreme left of $ (\spadesuit) $, we have a positive element, while on the extreme right of $ (\spadesuit) $, we have a negative element. This can only mean that both extremes are zero, so $ (p_{i} p_{j}) (p_{i} p_{j})^{*} = 0_{A} $. Hence,
$$
\| p_{i} p_{j} \|_{A}^{2}
= \| (p_{i} p_{j}) (p_{i} p_{j})^{*} \|_{A}
= \| 0_{A} \|_{A}
= 0,
$$
or equivalently, $ p_{i} p_{j} = 0_{A} $.