When discussing Hodge numbers, the identity $h^{p,q} = h^{n-p,n-q}$ is called Serre duality. AFAIU, this identification follows from applying the Hodge $*$ isomorphism. In particular, $H^q(X,\Omega^p_X)$ is dual to $H^{n-q}(X,\Omega^{n-p}_X)$.
If $X$ is now projective, we know that $H^q(X,\Omega_X^p)$ is dual to $H^{n-q}(X, \Omega_X^n \otimes \wedge^p T_X )$. I would like to then deduce the identity of Hodge numbers above by saying that $\Omega_X^{n-p}$ is isomorphic to $\Omega_X^n \otimes \wedge^p T_X$ but I don't know why that should be true.
It would certainly follow if I knew that the wedging map $\Omega^p_X \otimes \Omega^{n-p}_X \to \Omega^n_X$ were an isomorphism.
The result you're looking for is a special case of the following: "If $\mathcal F$ is a vector bundle of rank $n$, then $\wedge^p \mathcal F \cong (\wedge^{n-p} \mathcal F)^\vee\otimes \wedge^n \mathcal F$." (This is Hartshorne exercise 5.16.)
Note that $(\wedge^{n-p} \mathcal F)^\vee\otimes \wedge^n \mathcal F \cong \mathcal {Hom}_{\mathcal O}(\wedge^{n-p} \mathcal F, \wedge^n \mathcal F)$, so we just need to define an isomorphism from $\wedge^p \mathcal F$ to $\mathcal {Hom}_{\mathcal O}(\wedge^{n-p} \mathcal F, \wedge^n \mathcal F)$. To define this isomorphism, pick a trivialising open cover $\{ U_\alpha \}$. Since $\mathcal F|_{U_\alpha} \cong \mathcal O|_{U_\alpha}^{\oplus n}$, we can write any section of $\mathcal F|_{U_\alpha}$ as an $n$-component vector, whose components are sections of $\mathcal O$. We can then define our isomorphism from $\wedge^p \mathcal F|_{U_\alpha}$ to $\mathcal {Hom}_{\mathcal O|_{U_\alpha}}(\wedge^{n-p} \mathcal F|_{U_\alpha}, \wedge^n \mathcal F|_{U_\alpha})$ in the same way that we would define the natural isomorphism from $\wedge^p V $ to ${\rm Hom}(\wedge^{n-p}V, \wedge^n V)$ if $V$ is an $n$-dimensional vector space. Finally, this isomorphism is "basis independent", which ensures that it glues properly between the different $U_\alpha$'s.