So far as I can tell, it depends.

The map $\oplus BP_*(X_i) \to \prod BP_*(X_i)$ is injective and the cokernel $M$ is $BP_*$ of the cofiber. In particular, for any $N$ we have an isomorphism
$$
M \cong \prod_{i \geq 0} BP_*(X_i) \Big/ \bigoplus_{i \geq 0} BP_*(X_i) \cong \prod_{i \geq N} BP_*(X_i) \Big/ \bigoplus_{i \geq N} BP_*(X_i)
$$
As a result, properties of the modules $BP_*(X_i)$ that hold for all but finitely many exceptions tend to be inherited by the module $M$.

The first case is if the natural numbers $d_n$ go to $\infty$ as $n$ goes to $\infty$. In the identities $BP_*(X) = \oplus BP_* (X_i)$ and $BP_*(Y) = \prod BP_*(X_i)$, the direct sum and product are taken in the category of *graded* abelian groups. If the $d_n$ go to infinity, then for any fixed $k$ there are only finitely many $X_i$ with $BP_k(X_i) \neq 0$. As a result the group $M$ is zero in grading $k$. (The same is true for homotopy groups, and so the cofiber of the map $X \to Y$ is contractible). If you're asking for the $X_i$ to be suspension spectra, then I suspect that you're forced to have $d_n \to \infty$ (but I don't have a proof).

The second case is when the integers $d_n$ do not go to $\infty$. In this case, multiplication by $v_j^{k_j}$ acts by zero on all but finitely many $BP_*(X_i)$, and so the same is true for $M$. Thus means that $M$ consists entirely of things that are torsion.

What you might have had in mind was an "ultrafilter" type argument that did the exact opposite. If, instead, you have
$$
BP_*(X_i) \cong \Sigma^{d_i} BP_* / (v_0^{k_{0,i}}, v_1^{k_{1,i}}, \dots, v_i^{k_{i,i}})
$$
such that, for any fixed $n$, $k_{n,i} \to \infty$, then you have the opposite case: for all but finitely many $i$, $BP_m(X_i)$ has no $v_i^k$-torsion. As a result, the module $M$ has no torsion.

(You probably already know this, but I feel obligated to mention that the identity $BP_*(\prod X_i) \cong \prod BP_*(X_i)$ isn't always true, and relies on the $X_i$ being connective and $BP$ having ($p$-local) finite type.)