= Solution
With opposite magnetic moments, a fixed global <spin> is the low-resistivity species in one ferromagnetic layer and the high-resistivity species in the other. Thus both spin channels have the same four-layer averaged resistivity
$$
\bar\rho_{AP}=\frac{t_1\rho_{\rm FM}^L+t_1\rho_{\rm FM}^H+2t_2\rho_{\rm NM}}{2D}=\frac{a+b}{2D}.
$$
Each channel's square resistance is $(a+b)/(4D^2)$, and the two channels run in parallel. Therefore
$$
\boxed{\frac1{R_{\uparrow,\downarrow}}=\frac{8D^2}{a+b}.}
$$
Both channels encounter the strongly scattering magnetic layer; neither retains the uniformly favourable path available for parallel magnetizations.
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