Let $M$ be a von Neumann algebra and $NS(M)$ be its normal state space.
Set $P_0(M):=\{s_{\phi}: \phi \in NS(M)\}$. There is a conclusion: For any projection $p\in M$, it follows from Zorn's lemma that there is an orthogonal family $(P_i)_{i\in I}$ in $P_0(M)$ satisfying $p=\sum_{i\in I} P_i$. Where $s_{\phi}$ is the support projection of the normal state $\phi$.
I tried to construct a partial orederd set as following :
Set $S_p=\{J：p=\sum_{j\in J}P_j, P_j\in P_0(M), P_iP_j=0 \quad \text{for any }i\neq j\}$, $S_p$ is ordered by inclusion. But I cannot check every chain in $S_p$ has a upper bound in $S_p$.
Consider the following set: $$S_p:=\left\{ J \subseteq P_0(M)\ \middle|\ \sum_{P\in J}P \le p,\quad P\cdot Q = 0 \text{ for all $P,Q$ in $J$} \right\}$$ ie $S_p$ consists of all sets of mutually orthogonal support projections that sum up to something less than $p$ (this sum is to be understood to converge in the ultra-weak topology or in SOT or in whatever von Neumann topology you like to look at). Give this an order relation $$J\leq I :\iff \sum_{P\in J} P \leq \sum_{P\in I} P,$$ ie $J$ is less than $I$ if the sum over $I$ is closer to the upper bound $p$.
Now for any chain $J_i$ in $S_p$ define $J:=\bigcup_i J_i$.All projections in $J$ will remain mutually orthogonal and their sum will remain less than the upper bound $p$. Hence $J\in S_p$ and obviously $J\ge J_i$ for all $i$. By Zorn's Lemma there are maximal elements of $S_p$.
Whats left is to check that any maximal element of $S_p$ must necessarily sum up to $p$. Suppose that $J$ is a maximal element and denote with $P_J=\sum_{P\in J}P$. Then $p-P_J$ is an orthogonal projection, if it isn't $=0$ then there must be some support projection $P'\leq p-P_J$, you can then check that $J\cup\{P'\}\geq J$, contradicting maximality.