= Solution
The isothermal volume <Isothermal compressibility>[compressibility] is $\kappa=-(1/V)(\partial V/\partial p)_T$. For isotropic cubic compression $V\propto a^3$ and $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$, so
$$
\frac1{d_{hkl}}\left(\frac{\partial d_{hkl}}{\partial p}\right)_T
=\frac1a\left(\frac{\partial a}{\partial p}\right)_T=-\frac\kappa3.
$$
<Derivative>[Differentiate] <Bragg's law> $2d_{hkl}\sin\theta=\lambda$ at fixed incident <wavelength> and fixed reflection:
$$
0=\sin\theta\frac{\partial d}{\partial p}+d\cos\theta\frac{\partial\theta}{\partial p},
\qquad\boxed{\left(\frac{\partial\theta}{\partial p}\right)_T=\frac\kappa3\tan\theta,\quad\alpha=\frac13.}
$$
This <derivative> uses radians. It remains locally valid with a pressure-dependent <Isothermal compressibility>[compressibility], provided the cubic isotropic phase is retained. For a constant stipulated $\kappa$, integration gives
$$
\boxed{a(p)=a_0e^{-\kappa p/3},\qquad
\sin\theta(p)=\sin\theta_0e^{\kappa p/3}.}
$$
Thus $a_0(1-\kappa p/3)$ is only the first-order approximation, not the exact consequence of a constant logarithmic <Isothermal compressibility>[compressibility]. The finite-pressure calculation also assumes that the reflection still exists, the <crystal> remains in the same structure and $\sin\theta_0e^{\kappa p/3}\leq1$. These limitations matter for a large extrapolation.
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