(As far as I can tell, Sabrina's answer is wrong---see my comment there---. Here is a different argument)

$\def\cA{ A}\def\cC{B(H)}\def\cB{B}$ Let $\cA_\varphi =C^*(\varphi(\cA))\subset\cC$ and $\cB_\psi=C^*(\psi(\cB))\subset\cC$. By hypothesis these subalgebras commute with each other. By Theorem 3.5.3 in [Brown-Ozawa 2008] the map $\varphi\otimes\psi:\cA\odot\cB\to\cA_\varphi \odot\cB_\psi$ is completely positive and max-max bounded. Let $\rho:\cA_\varphi \odot\cB_\psi\to\cC$ be the $*$-homomorphism (here is where we use that $\cA_\varphi $ and $\cB_\psi$ commute) induced by $\rho(a\otimes b)=ab$. As $\rho$ is a representation of $\cA\otimes\cB$, we have that $\|\rho(x)\|\leq\|x\|_{\max}$ for all $x\in\cA\odot\cB$ by definition of the max norm. Now $$ \|(\varphi\times\psi)(x)\| =\|(\rho\circ(\varphi\otimes_{\max}\psi))(x)\|\leq\|(\varphi\otimes_{\max}\psi)(x)\|_{\max} \leq\|x\|_{\max}. $$

The complete positivity follows from the fact that we have written $\varphi\times\psi$ as a composition of completely positive maps.

(As far as I can tell, Sabrina's answer is wrong---see my comment there---. Here is a different argument)

$\def\cA{ A}\def\cC{B(H)}\def\cB{B}$ Let $\cA_\varphi =C^*(\varphi(\cA))\subset\cC$ and $\cB_\psi=C^*(\psi(\cB))\subset\cC$. By hypothesis these subalgebras commute with each other. By Theorem 3.5.3 in [Brown-Ozawa 2008, the same section where the exercise belongs] the map $\varphi\otimes\psi:\cA\odot\cB\to\cA_\varphi \odot\cB_\psi$ is completely positive and max-max bounded. Let $\rho:\cA_\varphi \odot\cB_\psi\to\cC$ be the $*$-homomorphism (here is where we use that $\cA_\varphi $ and $\cB_\psi$ commute) induced by $\rho(a\otimes b)=ab$. As $\rho$ is a representation of $\cA\otimes\cB$, we have that $\|\rho(x)\|\leq\|x\|_{\max}$ for all $x\in\cA\odot\cB$ by definition of the max norm. Now $$ \|(\varphi\times\psi)(x)\| =\|(\rho\circ(\varphi\otimes_{\max}\psi))(x)\|\leq\|(\varphi\otimes_{\max}\psi)(x)\|_{\max} \leq\|x\|_{\max}. $$

The complete positivity follows from the fact that we have written $\varphi\times\psi$ as a composition of completely positive maps.

(I have to admit that I cannot justify every step in Sabrina's answer---see my comment there---. Here is a different argument)

$\def\cA{ A}\def\cC{B(H)}\def\cB{B}$ Let $\cA_\varphi =C^*(\varphi(\cA))\subset\cC$ and $\cB_\psi=C^*(\psi(\cB))\subset\cC$. By hypothesis these subalgebras commute with each other. By Theorem 3.5.3 in [Brown-Ozawa 2008] the map $\varphi\otimes\psi:\cA\odot\cB\to\cA_\varphi \odot\cB_\psi$ is completely positive and max-max bounded. Let $\rho:\cA_\varphi \odot\cB_\psi\to\cC$ be the $*$-homomorphism (here is where we use that $\cA_\varphi $ and $\cB_\psi$ commute) induced by $\rho(a\otimes b)=ab$. As $\rho$ is a representation of $\cA\otimes\cB$, we have that $\|\rho(x)\|\leq\|x\|_{\max}$ for all $x\in\cA\odot\cB$ by definition of the max norm. Now $$ \|(\varphi\times\psi)(x)\| =\|(\rho\circ(\varphi\otimes_{\max}\psi))(x)\|\leq\|(\varphi\otimes_{\max}\psi)(x)\|_{\max} \leq\|x\|_{\max}. $$

The complete positivity follows from the fact that we have written $\varphi\times\psi$ as a composition of completely positive maps.

(As far as I can tell, Sabrina's answer is wrong---see my comment there---. Here is a different argument)

$\def\cA{ A}\def\cC{B(H)}\def\cB{B}$ Let $\cA_\varphi =C^*(\varphi(\cA))\subset\cC$ and $\cB_\psi=C^*(\psi(\cB))\subset\cC$. By hypothesis these subalgebras commute with each other. By Theorem 3.5.3 in [Brown-Ozawa 2008] the map $\varphi\otimes\psi:\cA\odot\cB\to\cA_\varphi \odot\cB_\psi$ is completely positive and max-max bounded. Let $\rho:\cA_\varphi \odot\cB_\psi\to\cC$ be the $*$-homomorphism (here is where we use that $\cA_\varphi $ and $\cB_\psi$ commute) induced by $\rho(a\otimes b)=ab$. As $\rho$ is a representation of $\cA\otimes\cB$, we have that $\|\rho(x)\|\leq\|x\|_{\max}$ for all $x\in\cA\odot\cB$ by definition of the max norm. Now $$ \|(\varphi\times\psi)(x)\| =\|(\rho\circ(\varphi\otimes_{\max}\psi))(x)\|\leq\|(\varphi\otimes_{\max}\psi)(x)\|_{\max} \leq\|x\|_{\max}. $$

The complete positivity follows from the fact that we have written $\varphi\times\psi$ as a composition of completely positive maps.