Yes, this is possible, assuming that $A$ (or $B$) is nuclear. The same argument below (using that exact $C^\ast$-algebras are locally reflexive) also works if $A$ or $B$ is exact and the tensor product is the spatial (aka minimal) tensor product.

The role of nuclearity is that any two-sided closed ideal $J \subseteq A\otimes B$ is the closed linear span of all tensor products $I_A \otimes I_B$ of two-sided closed contained in $J$, see Corollary 9.4.6 in Brown and Ozawa's book. Now, let $J_A \subseteq A$ be the intersection of all two-sided, closed ideals $J \subseteq A$ such that $p \in J \otimes B$ and let $I:= \bigcap J \otimes B$ where the intersection is indexed by such $J$. Clearly $J_A \otimes B \subseteq I$. For the converse (which is not as trivial as it a priori looks), take $I_A \otimes I_B \subseteq I$ where $I_A$ and $I_B$ are two-sided closed ideals in $A$ and $B$ respectively. Then $I_A \subseteq J_A$ and $I_B \subseteq B$, so $I_A \otimes I_B \subseteq J_A \otimes B$. By nuclearity of $A$, $I$ is the closed linear span of all such $I_A\otimes I_B$, so $I = J_A \otimes B$. In particular, $p\in J_A \otimes B$.

Now, let $a_1,\dots, a_n \in J_A$ and $b_1,\dots, b_n \in B$ such that
\begin{equation}
\| \sum_{i=1}^n a_i \otimes b_i - p \| < 1/2.
\end{equation}
Let $\pi \colon A \to \mathcal B(\mathcal H)$ be an irreducible representation such that $(\pi \otimes \mathrm{id}_B)(p) = 0$. The image of $(\pi \otimes \mathrm{id}_B)$ is canonically $\pi(A) \otimes B$, so by nuclearity of $A$ (or $B$) and exactness of maximal tensor products, it follows that
\begin{equation}
0 \to (\ker \pi)\otimes B \to A \otimes B \to \pi(A) \otimes B \to 0
\end{equation}
is exact, so the kernel of $(\pi \otimes \mathrm{id}_B)$ is $(\ker \pi )\otimes B$. Hence $p \in (\ker \pi)\otimes B$ so $a_1,\dots, a_n \in J_A \subseteq \ker \pi$ by construction of $J_A$.

If $A$ or $B$ is exact, we get $\ker (\pi \otimes \mathrm{id}_B)$ either by using that $A$ is locally reflexive or $B$ is exact to get that
\begin{equation}
0 \to (\ker \pi) \otimes_{\min{}} B \to A \otimes_{\min{}} B \to \pi(A) \otimes_{\min{}} B \to 0
\end{equation}
is exact.