Let $\varphi$ be a normal positive linear functional on a von Neumann algebra $\mathcal{M}$. Let $p$ be the central support projection of $\varphi$. Define a linear functional $\psi$ on $p\mathcal{M}$ by $\psi(a)=\varphi(qa)$, where $q$ is a projection in the center of $\mathcal{M}$. Assume that $\psi$ is proportional to $\varphi$. I want to show that either $\psi=0$ or $\psi=\varphi$.
Since $\psi$ is proportional to $\varphi$, there exists $t\in \mathbb{C}$ such that $\psi(\cdot)=t\varphi(\cdot)$. Now, for $a=pq$, we obtain that $\psi(pq)=\varphi(pq)=t\varphi(pq)$. From this, we obtain that $\varphi(pq)(1-t)=0$. If $\varphi(pq)=0$, then using Cauchy-Schwarz inequality, we obtain that $$|\psi(a)|^2=|\varphi(qa)|^2=|\varphi(pqa)|^2\le \varphi(pqqp)\varphi(a^*a)=\varphi(pq)\varphi(a^*a)=0$$ On the other hand, if $t=1$, we obtain that $\varphi(pq)=\varphi(p)=1$ by taking $a=p$. Hence, Already $pq$ is a central projection and since $p$ is the smallest central projection with the property that $\varphi(p)=1$, we must have that $pq\ge p$. Therefore, $q \ge p$. Now, from the equality $\varphi(pq)=\varphi(p)=1$, we obtain that $$\varphi(p(q-p)p)=0\implies q=p.$$ Is this proof alright? Thanks for the help.
Your argument is fine, with the exception that you cannot conclude that $q=p$. For instance, you could start with $q=1$.
The argument can be made more compact, though.
You have that $\varphi(qa)=t\varphi(a)$ for all $a\in M$. So $$ \varphi((t-q)a)=0,\qquad a\in M. $$ Taking $a=t-q$ gives us $\varphi((t-q)^*(t-q))=0$, and this implies $(t-q)p=0$. If $qp=0$, then $\psi=0$. Otherwise, $qp=tp$ implies that $t=1$ (first $|t|=1$ by looking at the norms, and then $t>0$ by evaluating at states). Thus $qp=p$ and $p\leq q$, which gives $\psi=\varphi$.