The answer is "yes". The following ideas can be found in standard texts.
The functional $\omega:T\mapsto \langle T\xi,\eta\rangle$ is normal, so it has a polar decomposition: $\omega = v |\omega|$ where $v\in V$ is a partial isometry, and $|\omega|$ is positive. $v$ is uniquely defined by the extra condition that $v^*v=s(|\omega|)$ the support projection of $|\omega|$. So $v^*\omega = v^*v|\omega| = |\omega|$. However, obviously $v^*\omega=\langle\cdot v^*\xi,\eta\rangle$.
At this point I cheat, and invoke a result from Kadison+Ringrose, Vol II, Prop 7.3.12. This says that as $|\omega|=\langle\cdot v^*\xi,\eta\rangle$ is positive, we can find $\xi'\in H$ with $|\omega|=\langle\cdot \xi',\xi'\rangle$. Clearly $\| |\omega| \| = \|\xi'\|^2$. As $e=v^*v$ is the support projection of $|\omega|$ it is central, and so we see that $|\omega| = \langle \cdot \xi',\xi'\rangle
= \langle \cdot e\xi',e\xi'\rangle$. It follows that $\| |\omega| \| = \|\xi'\|^2
= \|e\xi'\|^2$ and so $e\xi'=\xi'$. So also $\|v\xi'\|=\|\xi'\|$.
Now, $\omega = \langle \cdot v\xi',\xi'\rangle$, and so $\|\omega\|
= \||\omega|\| = \|\xi'\|^2 = \|v\xi'\| \|\xi'\|$.
So your formula for the norm is correct, and actually the infimum is obtained.
I will admit that this was harder than I thought. I wonder if anyone else has an
Edit: A general comment. Very often, we work with von Neumann algebras in "standard position". From your
bio I see you are interested in abstract harmonic analysis, so perhaps interested
in group von Neumann algebras $VN(G)$. Acting on $L^2(G)$ these are in standard
position. Then every normal functional arises as $\langle \cdot \xi,\eta \rangle$
for some $\xi,\eta$. So in this case, I wouldn't need to use the reference, and
the argument becomes a lot easier.