= Solution
In the printed convention $\rho$ is energy density, since the gravitational source is $8\pi G\rho/(3c^2)$. For nonrelativistic <pressureless matter>, $\rho_m=\rho_{m,0}(1+z)^3$. Define the <cosmological density parameters> by
$$
\Omega_{m,0}=\frac{8\pi G\rho_{m,0}}{3c^2H_0^2},
\qquad \Omega_{k,0}=-\frac{kc^2}{a_0^2H_0^2},
\qquad \Omega_{\Lambda,0}=\frac{\Lambda c^2}{3H_0^2}.
$$
The <Friedmann equation> then gives
$$
H(z)=H_0E(z),\qquad E(z)=\sqrt{\Omega_{m,0}(1+z)^3+
\Omega_{k,0}(1+z)^2+\Omega_{\Lambda,0}}.
$$
Let $n_c$ be the constant <comoving number density> of absorbers, and let $\sigma_p$ be their fixed proper interception area. Their proper <number density> is $n_p(z)=n_c(1+z)^3$, with $a_0=1$. The <redshift-time relation> gives a local proper photon-path increment
$$
d\ell=c|dt|=\frac{c\,dz}{(1+z)H(z)}.
$$
For an infinitesimal path, the expected number of interceptions, and to first order the probability of an interception, is $d\mathcal N=n_p\sigma_p\,d\ell$. Thus <absorber incidence and comoving number density> imply
$$
\boxed{\frac{d\mathcal N}{dz}
=\frac{n_c\sigma_pc}{H_0}\frac{(1+z)^2}{E(z)}.}
$$
Two factors survive: proper density brings three powers of $1+z$, whereas proper path length brings the inverse of one power. A fixed comoving area instead of the assumed fixed proper area would change the result. The formula is an incidence rate, not a probability bounded by one over a finite redshift interval; multiple interceptions can occur.
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