In bordism theory and algebraic topology, 4d spin bordism group is generated by $K3$ surface, while 4d $SO$ bordism group generated by $\mathbf{CP}^2$.

$K3$'s 4-manifold signature is $- 16$ and $\mathbf{CP}^2$'s 4-manifold signature is $+1$.

Are there ways that showing the 16 copies of $\mathbf{CP}^2$ is cobordant to the orientation reversing of $K3$? Namely, are there 5-manifold making these 17 disjoint pieces of manifolds (16 copies of $\mathbf{CP}^2$ and $\overline{K3}$) null bordant?

Relatedly $\mathbf{CP}^2$ is a non-spin manifold. The $K3$ is a spin manifold. Can 16 copies of $\mathbf{CP}^2$ be a spin manifold in some modified way? So that the spin version of the 16 copies of $\mathbf{CP}^2$ becomes cobordant to $\overline{K3}$?