Let $\mathcal C$ be an abelian category equipped with a closed symmetric monoidal structure. This implies in particular that the monoidal structure $\otimes$ is right exact in each variable. I care most about the situation where $\mathcal C$ is finite $\mathbb C$-linear in the sense of arXiv:1406.4204 (in which case the monoidal structure is closed iff it is right exact in each variable). Note that, unlike in that paper, I specifically care about monoidal structures which are not rigid. An example of the category I have in mind is the category $\mathrm{Mod}^f_A$ of finite-dimensional modules for any finite-dimensional commutative algebra $A$ (with $\otimes = \otimes_A$).
Recall that an object $P \in \mathcal C$ is projective if $\hom(P,-) : \mathcal C \to \mathrm{AbGp}$ is right exact. (It is already left exact.) Note that this has nothing to do with the monoidal structure.
An object $F \in \mathcal C$ is flat if $F \otimes : \mathcal C \to \mathcal C$ is left exact. (It is already right exact.) Note that this has everything to do with the monoidal structure.
Are projective objects necessarily flat?
Of course, in $\mathrm{Mod}_A^f$ they are. The other examples I usually use of non-rigid monoidal categories are the representation theories of non-Hopf bialgebras, but there every object is flat.
Best Answer
I believe the following is a counterexample. Let $\mathcal{A}$ and $\mathcal{B}$ be closed symmetric monoidal abelian categories such that the unit object $1\in\mathcal{B}$ is projective and let $F:\mathcal{A}\to\mathcal{B}$ be a non-exact strong symmetric monoidal functor which has a right adjoint $G:\mathcal{B}\to\mathcal{A}$. For instance, if $A$ is a commutative ring and $B$ is a commutative $A$-algebra which is not flat over $A$, you could have $\mathcal{A}=\mathrm{Mod}_A$ and $\mathcal{B}=\mathrm{Mod}_B$ and $F(M)=M\otimes_A B$. Let $\mathcal{C}=\mathcal{A}\times\mathcal{B}$, and equip it with the symmetric monoidal structure given by $$(M,V)\otimes (N,W)=(M\otimes N,F(M)\otimes W\oplus V\otimes F(N)\oplus V\otimes W).$$
The unit is $(1,0)$, and associativity follows from $F$ being strong symmetric monoidal. Furthermore, this monoidal structure is closed, with internal hom given by $$\operatorname{hom}((M,V),(N,W))=(\operatorname{hom}(M,N)\oplus G(\operatorname{hom}(V,W)),\operatorname{hom}(F(M),W)\oplus\operatorname{hom}(V,W)).$$
In this category, the object $(0,1)$ is projective by hypothesis, but it is not flat because $(M,0)\otimes (0,1)=(0,F(M))$ and $F$ is not exact.
In the example mentioned above where $\mathcal{A}=\mathrm{Mod}_A$ and $\mathcal{B}=\mathrm{Mod}_B$ and $B$ happens to be a quotient of $A$, this construction has the following intuitive explanation. The monoidal product is defined as if $(M,V)$ were secretly the $A$-module $M\oplus V$ and the tensor product is just the ordinary tensor product of $A$-modules. In particular, since $B=0\oplus B$ is not flat over $A$, the object $(0,B)$ is not flat. However, the category itself doesn't believe that $(M,V)$ is just a single $A$-module $M\oplus V$, and in particular the quotient map $A\to B$ does not exist as a map $(A,0)\to (0,B)$ that would cause $(0,B)$ to fail to be projective.
For a finite $\mathbb{C}$-linear version of this example, you can take $A$ and $B$ to be finite-dimensional $\mathbb{C}$-algebras and restrict to finitely generated modules everywhere.