Maybe this is a simple question, but I'm not sure about the following:
Let $\cal M$, $\cal N$ be von Neumann algebras and $X\subseteq \cal M$ a weakly dense (possibly separable) $C^*$-subalgebra. Let $\rho, \tau$ be tracial states on $\cal N$ where $\tau$ is normal and $\psi: \cal M \rightarrow \cal N$ a surjective, normal $*$-homomorphism.
The Question: Does $\rho (x) = \tau \circ \psi (x)$ for every $x\in X$ already imply that $\rho = \tau \circ \psi $?
I assume you mean that $\rho$ is a state on $\mathcal M$.
Unless you assume that $\rho$ is normal, the answer is no.
Take $\mathcal M=\mathcal N=\ell^\infty(\mathbb N)$, $X=c_0$, $\psi$ the identity map. Take $\tau=0$, and $\rho$ the limit along any ultrafilter $\omega$. Then $\tau$ is normal, $\tau=\rho$ on $X$, but $\tau\ne \rho$.