= Solution
For a centred <Gaussian random field>, <Wick theorem> expresses each even correlation function as a sum over all pairings of two-point functions; odd moments vanish. For example,
$$
\langle ABCD\rangle=\langle AB\rangle\langle CD\rangle+\langle AC\rangle\langle BD\rangle+\langle AD\rangle\langle BC\rangle.
$$
For six fields there are $5\cdot3\cdot1=15$ pairings. In a tree-level connected three-point calculation with one cubic vertex, each of the three external fields must contract with a different vertex field. This gives $3!=6$ connected <Wick contractions>. Pairings connecting two external fields instead are disconnected tadpole contributions, omitted from the <primordial bispectrum>.
Put $\mathcal F=(1-c_s^2)\mathcal A/c_s^2$. Using $dt=a\,d\tau$ and $\dot\zeta=\zeta'/a$, the integrated <interaction Hamiltonian> becomes
$$
\int dt\,H_{\rm int}(t)=-\int d\tau\,d^3x\,\frac{a\epsilon\mathcal F}{H}\zeta'^3
=\int d\tau\,d^3x\,\frac{\epsilon\mathcal F}{H^2\tau}\zeta'^3.
$$
The unequal-time <Wick contraction> required by the displayed operator order is
$$
\boxed{\langle\zeta(\mathbf k,0)\zeta'(\mathbf p,\tau)\rangle
=(2\pi)^3\delta^{(3)}(\mathbf k+\mathbf p)\,u_k(0)u_p'{}^*(\tau).}
$$
The conjugate is essential: the equal-time power spectrum alone does not specify the erroneous unstarred replacement printed in the target expression. The <in-in bispectrum conjugation rule> fixes the vacuum convergence and final sign.
Fourier transforming the vertex gives $(2\pi)^3\delta^{(3)}(\mathbf p_1+\mathbf p_2+\mathbf p_3)$ with three measures $d^3p_j/(2\pi)^3$. The six pairings supply the product of three external-vertex delta functions, each carrying $(2\pi)^3$. Integrating all three internal momenta leaves one overall $(2\pi)^3\delta^{(3)}(\mathbf k_1+\mathbf k_2+\mathbf k_3)$. Thus \b[the normalized connected contraction formula is]
$$
\boxed{\langle\zeta_{\mathbf k_1}\zeta_{\mathbf k_2}\zeta_{\mathbf k_3}\rangle_c
=(2\pi)^3\delta^{(3)}\!\left(\sum_i\mathbf k_i\right)
\operatorname{Re}\left[-12i\frac{\epsilon\mathcal F}{H^2}
\prod_i u_{k_i}(0)\int_{-\infty(1-i0)}^0\frac{d\tau}{\tau}\prod_i u_{k_i}'{}^*(\tau)\right].}
$$
This spells out what the printed permutations and delta functions must represent: all six permutations, full Fourier normalization, and conjugated vertex modes. Three cyclic permutations alone would miss a factor of two.
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