Let $(M,g)$ be an $n$-dimensional smooth Riemannian manifold without boundary. Given an endomorphism $A(x) : \sf T_x^*M \to \sf T_x^*M$ on the cotangent bundle, we can extend $A$ to an endomorphism on the $p$-th exterior power acting on a $p$-form $\alpha$ by \begin{align*} \wedge^p A : \wedge^p \sf T_x^*M &\to \wedge^p \sf T_x^*M\\ \wedge^p A(\alpha_1 \wedge ... \wedge \alpha_p) &:= A\alpha_1 \wedge ... \wedge A\alpha_p. \end{align*} How can we make sense of the codifferential $$\delta(\wedge^p A\alpha)?$$ The problem is that $\dim \wedge^p \sf T_x^*M = {n \choose p}$, unlike the case $\dim \wedge^n \sf T_x^*M = {n \choose n} = 1$, where $\wedge^n A$ must act as a multiplication operator, i.e. the determinant.
But $\delta$ is not a derivation of the algebra, i.e. there is no nice version of the Leibniz rule. Either using the definition of $\delta = \ast \, \mathbf d \, \ast $ directly or using the definition that it is the formal adjoint of the exterior derivative $\mathbf d$, the question boils down to: What is the exterior derivative of this $k$-th power endomorphism $\wedge^p A$?