 Solution

The coupling $\frac{c}{2}\int \varphi p^2$ is linear in $\varphi$, but its coefficient $p^2(\mathbf{r})$ is generally position dependent and dynamical. It therefore cannot be reduced to a constant times the conserved integral of $\varphi$; it changes both the local chemical potential of the mixture and the tendency of the polar order parameter to order.

Interchanging the labels of the two species sends $\varphi \mapsto -\varphi$ while leaving the even terms in $\varphi$ unchanged. It sends $c\mapsto -c$, so one may choose $c>0$ without loss of generality.

The isotropic disordered system is invariant under spatial inversion, which sends the polar order parameter $p\mapsto -p$ while leaving the scalar composition $\varphi$ unchanged. A term $\varphi p$ is odd under this symmetry and is forbidden; in more than one dimension it also fails to be a rotational scalar. Finally, if no higher even powers are retained, $b<0$ or $B<0$ makes the free energy unbounded below as the corresponding field grows. Thermodynamic stability therefore requires $b,B>0$.