On the page78 of the book C*-algebras and Finite-dimensional Approximations there is a corollary as follows.
Corollary 3.4.3. Let $\|\cdot\|_\alpha$ be a $C^\ast$-norm on $A\odot B$ and $\xi$ be a state on $A\otimes_\alpha B$. If $\xi|_A$ is pure, then $\xi|_{A\odot B}=\xi|_A\odot\xi|_B$.
Proof: Assume that $\xi|_A$ is pure. Applying the previous lemma to the vector state $T\mapsto\langle Tv_\xi,v_\xi\rangle$, it follows that $$\xi(a\otimes b)=\langle \pi_{\xi,A}(a)\pi_{\xi,B}(b)v_\xi,v_\xi\rangle\\ =\langle \pi_{\xi,A}(a)v_\xi,v_\xi\rangle\langle\pi_{\xi,B}(b)v_\xi,v_\xi\rangle$$
I can not understand why the second "=" hold. I think this need $\pi_{\xi,A}(a)\mapsto\langle \pi_{\xi,A}(a)v_\xi,v_\xi\rangle$ be a pure sate, but I can not get it from $\xi|_A$ is pure. Thenks a lot to anyone who give me a hint.
I guess the problem is that the notation $\xi|_A$ is used in two ways. In Lemma 3.4.2 is means the state $\xi$ restricted to $A$. But in Corollary 3.4.3, it means the state $a\mapsto\langle \pi_{\xi,A}(a)v_\xi,v_\xi\rangle$.
So the hypothesis is that this last state is pure. It is easy to see then that $\varphi:\pi_{\xi,A}(a)\mapsto\langle \pi_{\xi,A}(a)v_\xi,v_\xi\rangle$ is also pure.
Then you consider, for the purposes of Lemma 3.4.2, the algebras $\pi_{\xi,A}(A)\subset B(\mathcal H_\xi)$ and its commutant (that contains $\pi_{\xi,B}(B)$). So the proof goes $$ \xi(a\otimes b)=\langle \pi_\xi(a\otimes b)v_\xi,v_\xi\rangle=\langle \pi_{\xi,A}(a)\pi_{\xi,B}(b)v_\xi,v_\xi\rangle=\varphi(\pi_{\xi,A}(a)\pi_{\xi,B}(b))=\varphi(\pi_{\xi,A}(a))\varphi(\pi_{\xi,B}(b))=\langle \pi_{\xi,A}(a)v_\xi,v_\xi\rangle\,\langle \pi_{\xi,B}(b)v_\xi,v_\xi\rangle\\ =\xi|_A(a)\,\xi|_B(b)=(\xi|_A\odot\xi|_B)(a\otimes b) $$