Let $M$ be a finitely-generated projective module over a commutative algebra $A$, defined over $\mathbb{R}$. Consider a nondegenerate bilinear pairing $$ (*,*):M \times M \to A, $$ where by nondegenerate we mean that, for all non-zero $m \in M$, there exists some $n \in M$, such that $$ (m,n) \neq 0. $$
Does there exist an isomorphism $$ \phi:M \to M^* $$ such that $$ (m,n) = \phi(m)(n), $$ for all $m,n \in M$?
A side question, is: is projectivity necessary for us to establish such an isomorphism, or can it be established in a weaker setting?
No; for instance, if $A=M=\mathbb{R}[x]$ and $(m,n)=xmn$ then $\phi: M\to M^*\cong M$ is the map $\phi(m)=xm$ which is not an isomorphism.
More generally, $\phi(m)(n)=(m,n)$ defines a homomorphism $M\to M^*$ for any bilinear form on a module $M$, and nondegeneracy just says that $\phi$ is injective. In general there is no reason to think $\phi$ is surjective, though.