= Solution
Define the dimensionless parameter
$$
\delta=\left(\frac{\rho gh^4}{B}\right)^{1/4}.
$$
The stipulated hierarchy $kh\ll\delta\ll1$ implies both $kh\ll1$ and
$$
\frac{Bk^4}{\rho g}=\left(\frac{kh}{\delta}\right)^4\ll1.
$$
Thus the finite-depth correction uses $\tanh(kh)\sim kh$, while the restoring effect of <bending stiffness> is small compared with <gravity>. The <dispersion relation> becomes
$$
\omega^2
=ghk^2\left[1+\left(\frac{kh}{\delta}\right)^4\right]
\frac{\tanh(kh)}{kh}
\sim ghk^2.
$$
Hence, to leading order, these <surface gravity waves> are <nondispersive waves> with
$$
\boxed{\omega\sim k\sqrt{gh},\qquad c_p\sim c_g\sim c=\sqrt{gh}}.
$$
For clarity, the first corrections to the <phase speed> are
$$
\frac{c_p}{\sqrt{gh}}
=1+\frac12\left(\frac{kh}{\delta}\right)^4
-\frac16(kh)^2
+O\!\left((kh/\delta)^8+(kh)^4+(kh/\delta)^4(kh)^2\right).
$$
They exhibit the weak residual <wave dispersion> at finite <wavenumber>. The claimed nondispersive behavior is the asymptotic leading-order result, not exact equality for the complete relation.
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