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} $.