= Solution
First check the printed equations literally. Set $v=0$ and use $\rho_0'/\rho_0=-1/H$. Differentiating continuity in time and using horizontal momentum gives
$$
p_{xx}=\rho_0(w_{zt}-w_t/H).
$$
Vertical momentum and <buoyancy> give $w_{tt}+N^2w=-p_{zt}/\rho_0$. Combining them therefore yields
$$
\boxed{\partial_t^2\left(w_{xx}+w_{zz}-\frac2H w_z+\frac1{H^2}w\right)+N^2w_{xx}=0,}
$$
which differs from the requested equation. The factor two arises when differentiating both $\rho_0$ and $w_z-w/H$. Keeping variable <density> in continuity while treating it as constant during that differentiation would be inconsistent.
For the intended <anelastic reduced-pressure Kelvin wave>, make the explicit correction $w_t-b+\Phi_z=0$, while $u_t+\Phi_x=0$ and $\Phi=p'/\rho_0$. Continuity then gives $\Phi_{xx}=w_{zt}-w_t/H$, and vertical momentum gives $w_{tt}+N^2w=-\Phi_{zt}$. Cross differentiation now produces
$$
\boxed{\partial_t^2\left(w_{xx}+w_{zz}-\frac1H w_z\right)+N^2w_{xx}=0.}
$$
This proves the advertised equation for the corrected model, without claiming it follows from the incompatible printed set.
A concrete counterexample makes the distinction testable. In nondimensional units $H=N=k=1$, take $m=0$, $\omega=2/\sqrt5$ and $w=e^{z/2}e^{i(x-\omega t)}$, multiplied by its <Equatorial Kelvin wave>[Gaussian meridional envelope]. This satisfies the last boxed wave equation, but substitution into the literal printed-system wave operator gives $-2w/5\ne0$. No choice of accompanying <pressure> and horizontal velocity can then make it a solution of all five printed equations. Conversely, the literal operator has vertical factors $e^{z/H+imz}$ and dispersion $\omega^2=N^2k^2/(k^2+m^2)$; its scaling is different from the requested half-scale-height factor.
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