I think the solution is simply $$P=G_1 G_2 A^{-1}(1 +\lambda ).$$ Let me check by substitution $$G_1G_2P^\top(PAP^\top)^{-1}P=G_1 G_2(G_1G_2A^{-1})^\top\bigl(G_1G_2(G_1G_2A^{-1})^\top\bigr)^{-1}G_1G_2A^{-1}=G_1G_2A^{-1},$$ so yes, it solves $$P - G_1G_2(\lambda P^\top(PAP^\top)^{-1}P + A^{-1} ) = 0.$$

The solution for $P$ to $$P - G_1G_2(\lambda P^\top(PAP^\top)^{-1}P + A^{-1} ) = 0$$ is $$P=(1 +\lambda )G_1 G_2 A^{-1},$$ as one can check by substitution into $$G_1G_2P^\top(PAP^\top)^{-1}P=G_1 G_2(G_1G_2A^{-1})^\top\bigl(G_1G_2(G_1G_2A^{-1})^\top\bigr)^{-1}G_1G_2A^{-1}=G_1G_2A^{-1}.$$