Let $A$ and $B$ be $C^{\ast}$-algebras and $\pi: A \otimes B \to B(\mathcal{H}) $ be a representation of $A\otimes B$. Then there exists unique commuting representations $\pi_1$ of $A$ and $\pi_2$ of $B$ such that $$\pi(a \otimes b) = \pi_1(a) \pi_2(b) $$.
Suppose $\pi$ is $\textbf{irreducible}$ representation of $ A\otimes B$. Is it true that $\pi_1$ and $\pi_2$ will also be irreducible?
No. You will always have that $\pi(B)\subset\pi_1(A)'$, so a necessary condition for $\pi_1$ to be irreducible is that $B=\mathbb C$
Here's an example. Let $A=B=M_2(\mathbb C)$. Let $$ \pi_1(a)=\begin{bmatrix} a_{11}&a_{12}&0&0\\ a_{21}&a_{22}&0&0\\ 0&0&a_{11}&a_{12}\\ 0&0&a_{21}&a_{22}\\ \end{bmatrix} ,\qquad \pi_2(b)=\begin{bmatrix} b_{11}&0&b_{12}&0\\ 0&b_{11}&0&b_{12}\\ b_{21}&0&b_{22}&0\\ 0&b_{21}&0&b_{22} \end{bmatrix} $$ Then $\pi:A\otimes B\to M_4(\mathbb C)$ induced by $\pi(a\otimes b)=\pi_1(a)\pi_2(b)$ is an irreducible representation. And in this case $\pi_1(A)'=\pi_2(B)$, $\pi_2(B)'=\pi_1(A)$.