Let $D_A\subseteq A^*$ be the functionals of the form $\mu(\pi(\cdot)x)$, for $x\in X, \mu\in X^*, \|x\|\leq 1, \|\mu\|\leq 1$. As $\pi$ is an isometry, Hahn-Banach shows that the convex hull of $X$ is weak$^*$-dense in the closed ball of $A^*$, say $A^*_{[1]}$.

Similarly for $D_B$ using $\rho$. It's clear (*) that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in D_A, \mu_B\in D_B). \] However, this inequality is preserved by taking convex combinations of the elements of $D_A$ and $D_B$, and by taking weak$^*$-closures. But that would then show that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in A^*_{[1]}, \mu_B\in B^*_{[1]}), \] and that's all you need to show, I think?

Why is (*) true? Given $x\in X,\mu\in X^*,y\in Y,\lambda\in Y^*$, we have that $x\otimes y\in X\widehat\otimes Y$ of norm $\|x\|\|y\|$, and also $\mu\otimes\lambda \in (X\widehat\otimes Y)^*$ (this is $w\otimes z\mapsto \mu(w)\lambda(z)$) has norm $\|\mu\| \|\lambda\|$. Then $(\mu_A\otimes\mu_B)\tau = (\mu\otimes\lambda)((\pi\otimes\rho)(\tau)(x\otimes y))$. Actually, this shows that the result would remain true if you replaced the projective tensor norm on $X\otimes Y$ be any reasonable cross-norm.

Let $D_A\subseteq A^*$ be the functionals of the form $\mu(\pi(\cdot)x)$, for $x\in X, \mu\in X^*, \|x\|\leq 1, \|\mu\|\leq 1$. As $\pi$ is an isometry, Hahn-Banach shows that the convex hull of $X$ is weak$^*$-dense in the closed ball of $A^*$, say $A^*_{[1]}$.

Similarly for $D_B$ using $\rho$. It's clear (*) that

$$ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in D_A, \mu_B\in D_B). $$ However, this inequality is preserved by taking convex combinations of the elements of $D_A$ and $D_B$, and by taking weak$^*$-closures. But that would then show that $$ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in A^*_{[1]}, \mu_B\in B^*_{[1]}), $$ and that's all you need to show, I think?

Why is (*) true? Given $x\in X,\mu\in X^*,y\in Y,\lambda\in Y^*$, we have that $x\otimes y\in X\widehat\otimes Y$ of norm $\|x\|\|y\|$, and also $\mu\otimes\lambda \in (X\widehat\otimes Y)^*$ (this is $w\otimes z\mapsto \mu(w)\lambda(z)$) has norm $\|\mu\| \|\lambda\|$. Then $(\mu_A\otimes\mu_B)\tau = (\mu\otimes\lambda)((\pi\otimes\rho)(\tau)(x\otimes y))$. Actually, this shows that the result would remain true if you replaced the projective tensor norm on $X\otimes Y$ be any reasonable cross-norm.

Let $D_A\subseteq A^*$ be the functionals of the form $\mu(\pi(\cdot)x)$, for $x\in X, \mu\in X^*, \|x\|\leq 1, \|\mu\|\leq 1$. Hahn-Banach shows that the convex hull of $X$ is weak$^*$-dense in the closed ball of $A^*$, say $A^*_{[1]}$.

Similarly for $D_B$ using $\rho$. It's clear (*) that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in D_A, \mu_B\in D_B). \] However, this inequality is preserved by taking convex combinations of the elements of $D_A$ and $D_B$, and by taking weak$^*$-closures. But that would then show that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in A^*_{[1]}, \mu_B\in B^*_{[1]}), \] and that's all you need to show, I think?

Why is (*) true? Given $x\in X,\mu\in X^*,y\in Y,\lambda\in Y^*$, we have that $x\otimes y\in X\widehat\otimes Y$ of norm $\|x\|\|y\|$, and also $\mu\otimes\lambda \in (X\widehat\otimes Y)^*$ (this is $w\otimes z\mapsto \mu(w)\lambda(z)$) has norm $\|\mu\| \|\lambda\|$. Then $(\mu_A\otimes\mu_B)\tau = (\mu\otimes\lambda)((\pi\otimes\rho)(\tau)(x\otimes y))$. Actually, this shows that the result would remain true if you replaced the projective tensor norm on $X\otimes Y$ be any reasonable cross-norm.

Let $D_A\subseteq A^*$ be the functionals of the form $\mu(\pi(\cdot)x)$, for $x\in X, \mu\in X^*, \|x\|\leq 1, \|\mu\|\leq 1$. As $\pi$ is an isometry, Hahn-Banach shows that the convex hull of $X$ is weak$^*$-dense in the closed ball of $A^*$, say $A^*_{[1]}$.

Similarly for $D_B$ using $\rho$. It's clear (*) that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in D_A, \mu_B\in D_B). \] However, this inequality is preserved by taking convex combinations of the elements of $D_A$ and $D_B$, and by taking weak$^*$-closures. But that would then show that \[ |(\mu_A\otimes\mu_B)\tau| \leq \| (\pi\otimes\rho)(\tau) \|_{op} \qquad (\tau\in A\otimes B, \mu_A\in A^*_{[1]}, \mu_B\in B^*_{[1]}), \] and that's all you need to show, I think?

Why is (*) true? Given $x\in X,\mu\in X^*,y\in Y,\lambda\in Y^*$, we have that $x\otimes y\in X\widehat\otimes Y$ of norm $\|x\|\|y\|$, and also $\mu\otimes\lambda \in (X\widehat\otimes Y)^*$ (this is $w\otimes z\mapsto \mu(w)\lambda(z)$) has norm $\|\mu\| \|\lambda\|$. Then $(\mu_A\otimes\mu_B)\tau = (\mu\otimes\lambda)((\pi\otimes\rho)(\tau)(x\otimes y))$. Actually, this shows that the result would remain true if you replaced the projective tensor norm on $X\otimes Y$ be any reasonable cross-norm.