= Solution
Put $k_1=q\ll k$ and $k_2,k_3\to k$ in the <squeezed bispectrum configuration>. Then $K\to2k$ and the three symmetric combinations in the supplied expression have limits
$$
\sum_i k_i^3\to2k^3,\qquad\sum_{i\ne j}k_i k_j^2\to2k^3,\qquad\sum_{i>j}k_i^2k_j^2\to k^4.
$$
The first two terms inside the $\epsilon$ bracket cancel, leaving $4k^3$. The full bracket is therefore $(\eta+2\epsilon)k^3/4$, and
$$
\boxed{F(q,k,k)\simeq(2\pi)^3\frac{H^4}{16\epsilon^2M_p^4}\frac{\eta+2\epsilon}{q^3k^3}.}
$$
The $(2\pi)^3$ is kept in the normalization of $F$ used by the paper; it should not be introduced a second time by changing the <primordial bispectrum> convention.
<Momentum conservation> makes the three vectors close into a triangle. As $q/k\to0$, the two short-wavelength vectors approach equal magnitude and opposite direction, producing a thin, nearly collapsed triangle. In real space the $q$ mode varies slowly over a region containing many short wavelengths, so it modulates their local statistics. For a single adiabatic inflationary clock this is a local spatial dilation, the basis of the <single-field inflation squeezed-limit consistency relation>. If $\eta$ is the Hubble-flow convention $\dot\epsilon/(H\epsilon)$, the <scalar spectral index in Hubble slow-roll parameters> gives $\eta+2\epsilon=1-n_s$ to first order. This identification must not be made with the potential slow-roll $\eta$ without converting conventions.
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