Does every enriched functor preserve tensors?

No. The functor “take a vector space to its double dual” is linear but does not preserve tensors with infinite-dimensional vector spaces.

Enriched functors are automatically “lax tensored”. An enriched functor $F$ provides a natural map $[X,Y] \overset F\to [FX,FY]$ for any $X,Y$. Consider setting $Y=P\otimes X$, and study:
$$\mathrm{id} \in [P\otimes X,P\otimes X] \cong [P,[X,P\otimes X]] \overset {F\circ} \to [P,[FX,F(P\otimes X)]] \cong [P\otimes FX,F(P\otimes X)].$$
This gives a canonical map $P\otimes FX \to F(P\otimes X)$ which is natural in $P,X$ and compatible with associativity an unit data. It just isn’t automatically an isomorphism.

On first reading, I missed the requirement that $\mathcal P$ be rigid. In that case the answer is Yes, enriched functors are automatically tensored. In general, a-priori-lax constructions become strong when there is enough dualizability. Consider the composition
$$F(P\otimes X) \to P \otimes P^* \otimes F(P\otimes X) \to P \otimes F(P^* \otimes P \otimes X) \to P \otimes F(X)$$
where the first map is the unit between $P,P^*$, the second is the lax monoidality constructed above, and the third is $F$ of the counit. This should give an inverse to the lax monoidality.