= Solution
At late time $f(k,\tau_0)\to H/\sqrt{2k^3}$ and $a=-1/(H\tau)$. Define the <elementary symmetric polynomials>
$$
e_2=\sum_{a<b}k_ak_b,
\qquad e_3=\sum_{a<b<c}k_ak_bk_c,
\qquad e_4=k_1k_2k_3k_4.
$$
Then part ii reduces to
$$
\langle O\rangle'_\lambda
=-\frac{i\lambda H^4\mathcal E}{8\prod_a k_a^3}\operatorname{Im}J(\tau_0),
$$
where a prime removes the momentum-conserving <Dirac delta distribution> and
$$
J(\tau_0)=\int_{-\infty(1-i\epsilon)}^{\tau_0}
\frac{d\tau}{\tau^4}e^{-ik_T\tau}
\prod_{a=1}^4(1+ik_a\tau).
$$
Expanding the product gives
$$
\frac1{\tau^4}\prod_a(1+ik_a\tau)
=\frac1{\tau^4}+\frac{ik_T}{\tau^3}-\frac{e_2}{\tau^2}
-\frac{ie_3}{\tau}+e_4.
$$
Repeated <integration by parts> reduces every negative power to the supplied logarithmic integral. The power divergences are real and disappear when the imaginary part is taken. Writing
$$
L_T=\gamma_E+\log|k_T\tau_0|,
$$
one obtains
$$
\operatorname{Im}J
=\left(-\frac{k_T^3}{3}+k_Te_2-e_3\right)L_T
+\frac{4k_T^3}{9}-k_Te_2+\frac{e_4}{k_T}+o(1).
$$
Consequently the late-time <parity-odd primordial trispectrum> is
$$
\boxed{
\langle\phi_1(\mathbf k_1)\phi_2(\mathbf k_2)
\phi_3(\mathbf k_3)\phi_4(\mathbf k_4)\rangle'
=-\frac{i\lambda H^4}{8\prod_a k_a^3}
[\mathbf k_2\mathbin\cdot(\mathbf k_3\mathbin\times\mathbf k_4)]
\left[
\left(-\frac{k_T^3}{3}+k_Te_2-e_3\right)L_T
+\frac{4k_T^3}{9}-k_Te_2+\frac{e_4}{k_T}
\right]}.
$$
Its factor of $i$ is required by <reality of a momentum-space scalar correlator>: reversing all momenta complex-conjugates the correlator, while the <scalar triple product> changes sign.
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