[proofplan]
We construct $c(p)$ as the meet, inside the projection lattice of the center $Z(M)$, of all central projections that dominate $p$. Completeness of the projection lattice gives existence, and the meet property gives both domination of $p$ and minimality among central majorants. The annihilator identity then follows from the elementary projection fact that, for commuting projections, $ab=0$ is equivalent to $a \le 1-b$.
[/proofplan]
[step:Construct the central carrier as a meet of central majorants]
Let $\mathcal{S}$ denote the set of central projections in $M$ that dominate $p$:
\begin{align*}
\mathcal{S}:=\{q\in \mathcal{P}(Z(M)):p\le q\}.
\end{align*}
The set $\mathcal{S}$ is nonempty because $1\in \mathcal{S}$. Since $Z(M)$ is a von Neumann algebra, [citetheorem:9267] applied to $Z(M)$ shows that $\mathcal{P}(Z(M))$ is a complete lattice. Therefore the meet
\begin{align*}
c(p):=\bigwedge_{q\in\mathcal{S}} q
\end{align*}
exists in $\mathcal{P}(Z(M))$. Hence $c(p)$ is a central projection.
For every $q\in\mathcal{S}$, the inequality $p\le q$ means $pH\subseteq qH$. By the range formula for meets in [citetheorem:9267], applied in $Z(M)$,
\begin{align*}
c(p)H=\bigcap_{q\in\mathcal{S}}qH.
\end{align*}
Thus $pH\subseteq c(p)H$, so $p\le c(p)$. Also, by definition of meet, $c(p)\le q$ for every $q\in\mathcal{S}$. Therefore $c(p)$ is the smallest central projection dominating $p$.
[guided]
The goal is to manufacture the smallest central projection above $p$. The natural way to do this is to take all possible central projections above $p$ and then take their greatest lower bound.
Define
\begin{align*}
\mathcal{S}:=\{q\in \mathcal{P}(Z(M)):p\le q\}.
\end{align*}
This is a set of projections in the center $Z(M)$. It is nonempty because the identity projection $1$ is central and satisfies $p\le 1$. Since $Z(M)$ is itself a von Neumann algebra, [citetheorem:9267] applies to $Z(M)$ and says that its projection lattice is complete. Therefore the meet
\begin{align*}
c(p):=\bigwedge_{q\in\mathcal{S}}q
\end{align*}
exists as a projection in $Z(M)$. In particular, $c(p)$ is central.
We still have to check that this meet actually dominates $p$. For each $q\in\mathcal{S}$, the relation $p\le q$ for projections means that the range of $p$ is contained in the range of $q$, so $pH\subseteq qH$. Hence
\begin{align*}
pH\subseteq \bigcap_{q\in\mathcal{S}}qH.
\end{align*}
By the range description of meets in [citetheorem:9267], the right-hand side is exactly $c(p)H$. Thus $pH\subseteq c(p)H$, which is equivalent to $p\le c(p)$.
Finally, because $c(p)$ is the meet of all elements of $\mathcal{S}$, it satisfies $c(p)\le q$ for every $q\in\mathcal{S}$. Therefore any central projection that dominates $p$ also dominates $c(p)$. This proves that $c(p)$ is the smallest central projection dominating $p$.
[/guided]
[/step]
[step:Convert vanishing products of commuting projections into order relations]
We record the projection-order fact used below. Let $e$ and $f$ be commuting projections in $M$. Then
\begin{align*}
ef=0 \quad \Longleftrightarrow \quad e\le 1-f.
\end{align*}
Indeed, if $ef=0$, then commutativity gives $fe=0$, and hence $(1-f)e=e-fe=e$. Therefore, for every $\xi\in H$, one has $e\xi=(1-f)e\xi\in (1-f)H$, so $eH\subseteq (1-f)H$ and hence $e\le 1-f$. Conversely, if $e\le 1-f$, then $eH\subseteq (1-f)H$, so $f$ vanishes on $eH$, which gives $fe=0$. Since $e$ and $f$ commute, $ef=fe=0$.
[/step]
[step:Prove the annihilator identity for central projections]
Let $z\in Z(M)$ be a projection. Suppose first that $zc(p)=0$. Since $p\le c(p)$, we have $pH\subseteq c(p)H$. The equality $zc(p)=0$ means that $z$ vanishes on $c(p)H$, and therefore $z$ vanishes on $pH$. Hence $zp=0$.
Conversely, suppose that $zp=0$. Since $z$ is central, $z$ commutes with $p$, so the projection-order fact gives $p\le 1-z$. The projection $1-z$ is central, and it dominates $p$, so $1-z\in\mathcal{S}$. By the minimality proved above,
\begin{align*}
c(p)\le 1-z.
\end{align*}
Applying the projection-order fact again to the commuting projections $z$ and $c(p)$ gives
\begin{align*}
zc(p)=0.
\end{align*}
Thus $zp=0$ if and only if $zc(p)=0$ for every central projection $z\in Z(M)$.
[/step]
