= Solution
Let the common uncoupled <Fermi wavevector> be $k_0$. Equal <carrier densities> imply equal <Fermi surface> volumes, and hence equal radii for the spherical spin-degenerate electron pockets. The conserved total density is
$$
n_{\rm tot}=\frac{2k_0^3}{3\pi^2}.
$$
If hybridisation leaves only one occupied electron-like band, state counting gives
$$
\boxed{k_F=2^{1/3}k_0,\qquad p_F=2^{1/3}p_0,\qquad
\frac{\Delta p_F}{p_0}=2^{1/3}-1\simeq0.260.}
$$
This is the familiar single-large-Fermi-surface answer. It requires the other hybridised band to be empty.
\b[For finite heavy-band mass, that conclusion does not follow from an arbitrary nonzero coupling.] At sufficiently small $\alpha$, the upper band still has an occupied pocket, so there are two Fermi momenta, not one. The exact fixed-density statement is <Fermi-volume conservation in two hybridized electron bands>:
$$
\boxed{k_{F,-}^3+k_{F,+}^3=2k_0^3.}
$$
For an explicit weak-coupling check, write $E_p=Ak^2$, $E_h=C+Bk^2$, with $0<B\ll A$, and enforce the uncoupled equality at $k_0$: $C=(A-B)k_0^2$, $\mu_0=Ak_0^2$. Put $\mu=\mu_0+\delta\mu$ and $y=k_F^2-k_0^2$. The Fermi-surface equation is
$$
(\delta\mu-Ay)(\delta\mu-By)-\alpha^2=0.
$$
At fixed density $\delta\mu=O(\alpha^2)$, giving
$$
\boxed{k_{F,\mp}=k_0\pm\frac{\alpha}{2k_0\sqrt{AB}}+O(\alpha^2).}
$$
For arbitrary coupling, put $q_\pm=k_{F,\mp}^2$. The roots of the same <characteristic equation>, before discarding unoccupied pockets, are
$$
q_\pm=\frac{(A+B)\mu-AC\pm\sqrt{[(A-B)\mu-AC]^2+4AB\alpha^2}}{2AB}.
$$
Choose the <chemical potential> so that $[\max(q_+,0)]^{3/2}+[\max(q_-,0)]^{3/2}=2k_0^3$. This specifies the <Fermi momentum> of each pocket for any supplied curvatures and coupling.
The transition to one occupied pocket occurs at $\mu=E_+(0)$. Set $q_*=2^{2/3}k_0^2$; at the transition the two squared radii are $q_*$ and zero. Their sum and product give
$$
\mu_c=\frac{A(C+Bq_*)}{A+B},\qquad
\boxed{\alpha_c^2=\mu_c(\mu_c-C)
=\frac{AB(C+Bq_*)(Aq_*-C)}{(A+B)^2}.}
$$
For $0<\alpha<\alpha_c$ there are two pockets; for $\alpha\geq\alpha_c$ the only nonzero radius is $2^{1/3}k_0$. The outer radius grows continuously and the inner one shrinks. The nearly <flat band> limit can make depopulation occur at a small coupling, but the mass ratio or a single-surface assumption is needed to turn the question into a unique finite-coupling numerical change.
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