I believe the following is a counterexample. Let $A$ and $B$ be commutative rings and let $A\to B$ be an epimorphism such that $B$ is not flat over $A$. Let $\mathcal{C}=\mathrm{Mod}_A\times \mathrm{Mod}_B$ (also known as $\mathrm{Mod}_{A\times B}$), and equip it with the symmetric monoidal structure given by $$(M,V)\otimes (N,W)=(M\otimes N, V\otimes W\oplus M\otimes W\oplus V\otimes N),$$ where on the right-hand side $\otimes$ denotes the usual tensor product of $A$-modules (considering $B$-modules as $A$-modules via our map $A\to B$). This structure is closed, with internal hom given by $$\operatorname{hom}((M,V),(N,W))=(\operatorname{Hom}(M,N)\oplus\operatorname{Hom}(V,W),\operatorname{Hom}(M\otimes B,W)\oplus\operatorname{Hom}(V,W)),$$ where again on the right-hand side we take ordinary Homs of $A$-modules. In this category, the object $(0,B)$ is projective but not flat. Morally, what seems to be going on here is that the monoidal structure believes that $(M,V)$ is secretly the $A$-module $M\oplus V$, but the category itself doesn't know this (and in particular doesn't have any maps from $(A,0)$ to $(0,B)$ that would break the projectivity of $(0,B)$).
For a finite $\mathbb{C}$-linear version of this example, you can take $B=\mathbb{C}$ and $A$ to be an augmented finite-dimensional $\mathbb{C}$-algebra and restrict to finitely generated modules everywhere.