= Solution
Translate the concentric spheres so that their centre is $(0,0,-a)$, take $b=a+d$, and keep the <point charge> fixed at $(0,0,h)$. Its distance from the centre is $r_0=a+h$. As $a\to\infty$, the upper portions of the inner and outer spheres approach the planes $z=0$ and $z=d$. The preceding total induced <electric charge> becomes
$$
Q_a=-q\frac{a(d-h)}{(a+h)d}\longrightarrow-\frac{q(d-h)}d.
$$
The outer sphere has total induced <electric charge> $Q_b=-q-Q_a$: a <Gaussian surface> within its conducting material has zero <electric flux>, and the grounded exterior is field-free by <uniqueness of Poisson equation>. Therefore $Q_b\to-qh/d$.
One can justify the total-charge limit without assuming that local convergence of <electric fields> preserves their <surface integrals>. Use <induced charge as harmonic measure>: if a <harmonic function> $w$ is one on the selected grounded boundary and zero on the other, <Green's second identity> with the <electric potential> gives
$$
Q_{\rm selected}=-q\,w(\mathbf r_0).
$$
Indeed, $\nabla^2\phi=-q\delta^{(3)}(\mathbf r-\mathbf r_0)/\epsilon_0$, $\phi=0$ on the boundaries, and their <unit normal> outward from the gap satisfies $\sigma=\epsilon_0\partial_n\phi$. For the spheres the weighting <harmonic function> is $w_a(r)=a(b-r)/(r(b-a))$, exactly the factor already obtained. For the slab the two weighting <harmonic functions> are $w_0=1-z/d$ and $w_d=z/d$. Applying the same identity to the decaying grounded-slab <electric potential>, whose boundary term at lateral infinity vanishes, establishes directly
$$
\boxed{Q_0=-\frac{q(d-h)}d,\qquad Q_d=-\frac{qh}d.}
$$
Their sum is $-q$, and their relative sizes measure how close the <point charge> lies to each grounded plane.
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