Let $\alpha_1,\alpha_2,\alpha_3$ denote the dual basis associated with $v_1,v_2,v_3$. Note that $\alpha_1 \wedge \alpha_2, \alpha_1 \wedge \alpha_3, \alpha_2 \wedge \alpha_3$ is the dual basis associated with $v_1 \wedge v_2, v_1 \wedge v_3, v_2 \wedge v_3$, where we have
$$
(\alpha \wedge \beta)(v \wedge w) = \det \pmatrix{\alpha(v) & \alpha(v)\\ \beta(v) & \beta(w)}.
$$
Note that for a space $V$ with basis $B = (v_1,\dots,v_n)$ and dual basis $\alpha_1,\dots,\alpha_n$, the $i$th diagonal element of the matrix of $\phi:V \to V$ with respect to $B$ is $\alpha_i(\phi(v_i))$. With that established, note that
$$
(\alpha_i \wedge \alpha_j)(p\wedge p)(v_i \wedge v_j) =
(\alpha_i \wedge \alpha_j)(pv_i \wedge pv_j) = \\
\det \pmatrix{\alpha_i \circ pv_i & \alpha_i \circ pv_j\\
\alpha_j \circ pv_i & \alpha_j \circ pv_j}.
$$
Proof of the determinant formula:
Let $m_{ij} = \alpha_i (pv_j)$. We have $pv_j = \sum_i m_{ij}v_i$. Thus, we have
$$
\begin{align}
(p \wedge p)(v_i \wedge v_j) &=
pv_i \wedge pv_j
\\ & =
\left(\sum_{k} m_{ki}v_k \right) \wedge
\left(\sum_{\ell} m_{\ell j}v_\ell \right)
\\ & =
\sum_{k,\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
\\ & =
\sum_{k<\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
+ \sum_{k>\ell} -m_{ki}m_{\ell j} (v_\ell \wedge v_k)
\\ & =
\sum_{k<\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
+ \sum_{k'<\ell'} -m_{\ell'i}m_{k' j} (v_{k'} \wedge v_{\ell'})
\\ & =
\sum_{k<\ell} (m_{ki}m_{\ell j} - m_{kj}m_{\ell i}) (v_k \wedge v_\ell)
=
\sum_{k<\ell} \det\pmatrix{m_{ki} & m_{\ell i}\\ m_{kj} & m_{\ell j}} \cdot (v_k \wedge v_\ell).
\end{align}
$$
