= Solution
The magnetic bending angle is $\phi\simeq|q|BL/p$. For a <momentum> estimate based on this angle, <propagation of uncertainty> gives $\sigma(p)/p=\sigma(\phi)/\phi$. Comparing the total <Multiple Coulomb scattering> angle with the bend therefore gives the commonly intended characteristic estimate
$$
\boxed{\left(\frac{\sigma(p)}p\right)_{\rm ms}\sim
\frac{p\operatorname{RMS}(\theta)}{|q|BL}
=\frac{\sqrt{32\pi K\rho_nx\ln(\theta_{\max}/\theta_{\min})}}{|q|BL}.}
$$
Its independence of $p$ is the key result for relativistic particles: the scattering angle and magnetic bend both scale as $1/p$.
A precise standard deviation requires specifying how tracks are measured. If $\phi$ is obtained from independently measured entrance and exit directions and all the material lies between them, the bending-plane projected <root mean square> is $\operatorname{RMS}(\theta)/\sqrt2$. For that estimator the equality is
$$
\left(\frac{\sigma(p)}p\right)_{\rm ms,angle}
=\frac{\sqrt{16\pi K\rho_nx\ln(\theta_{\max}/\theta_{\min})}}{|q|BL}.
$$
A three-point <sagitta> fit weights scatter positions differently again. The paper does not specify these measurement details, so it fixes the scaling and characteristic estimate, but not one universal geometric coefficient for all track fits.
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