= Solution
For the <P(X, phi) scalar field theory>, the background pressure is $p=P$ and the <energy density> is $\rho=2XP_X-P$. Differentiating at fixed $\phi$ gives $p_X=P_X$ and $\rho_X=P_X+2XP_{XX}$, so the <Sound speed of a P(X, phi) scalar perturbation> is
$$
\boxed{c_s^2=\frac{p_X}{\rho_X}=\frac{P_X}{P_X+2XP_{XX}}}.
$$
This is the propagation speed of the scalar fluctuation; the derivative is taken at fixed field, rather than along an arbitrary background trajectory.
On the fixed expanding metric,
$$
X=\frac12\dot{\bar\phi}^{\,2}+\dot{\bar\phi}\,\delta\dot\phi+
\frac12(\delta\dot\phi)^2-\frac1{2a^2}(\partial_i\delta\phi)^2.
$$
Write the <comoving curvature perturbation> in the paper's convention as $\zeta=-H\delta\phi/\dot{\bar\phi}$. Neglecting <slow-roll> derivatives of $H$ and $\dot{\bar\phi}$ yields
$$
\boxed{\delta X=\frac{\bar X}{H^2}
\left[-2H\dot\zeta+\dot\zeta^2-a^{-2}(\partial_i\zeta)^2\right]}.
$$
With $\delta X_1=-2\bar X\dot\zeta/H$ and $\delta X_2=\bar X[\dot\zeta^2-a^{-2}(\partial_i\zeta)^2]/H^2$, the only cubic terms are
$$
P_{XX}\delta X_1\delta X_2+\frac16P_{XXX}\delta X_1^3
=\frac{2\bar X^2P_{XX}}{H^3}\dot\zeta\,a^{-2}(\partial_i\zeta)^2
-\left(\frac{2\bar X^2P_{XX}}{H^3}+
\frac{4\bar X^3P_{XXX}}{3H^3}\right)\dot\zeta^3.
$$
The identities $\epsilon=\bar XP_X/H^2$ and $(1-c_s^2)/c_s^2=2\bar XP_{XX}/P_X$ give the <cubic curvature action for a P(X, phi) scalar field>
$$
\boxed{
S_3=\int dt\,d^3x\,\frac{a^3\epsilon(1-c_s^2)}{Hc_s^2}
\left[a^{-2}\dot\zeta(\partial_i\zeta)^2+\mathcal A\dot\zeta^3\right],
\qquad
\mathcal A=-1-\frac{2\bar XP_{XXX}}{3P_{XX}}}.
$$
If $P_{XX}=0$, the unfactored expression above supplies the regular limiting result.
For the requested <primordial bispectrum>, define $C=\epsilon(1-c_s^2)/(Hc_s^2)$ and $\int_{\mathbf p}=\int d^3p/(2\pi)^3$. The cubic Hamiltonian and $dt=a\,d\tau$ combine to give
$$
aH_{\rm int}
=C a\int_{\mathbf p_1,\mathbf p_2,\mathbf p_3}
(2\pi)^3\delta^{(3)}(\mathbf p_1+\mathbf p_2+\mathbf p_3)
(\mathbf p_2\cdot\mathbf p_3)\,
\zeta'_{\mathbf p_1}\zeta_{\mathbf p_2}\zeta_{\mathbf p_3}.
$$
Here the positive sign comes from the two spatial derivatives and $H_{\rm int}=-L_{\rm int}$ at cubic order. <Wick theorem> gives two contractions for each choice of which external leg meets $\zeta'$. At the external time zero,
$$
\langle\zeta_{\mathbf k}(0)\zeta_{\mathbf p}(\tau)\rangle
=(2\pi)^3\delta^{(3)}(\mathbf k+\mathbf p)\,
u_k(0)u_k^*(\tau).
$$
Thus, after removing the momentum-conserving delta function, the Hamiltonian contraction is
$$
2Ca\prod_{j=1}^3u_{k_j}(0)
\sum_{\rm cyc}(\mathbf k_2\cdot\mathbf k_3)
u_{k_1}^{*\,\prime}(\tau)u_{k_2}^*(\tau)u_{k_3}^*(\tau).
$$
Put $A_k=H/\sqrt{4\epsilon c_sk^3}$. Then $u_k(0)=A_k$ and $u_k^{*\,\prime}=A_kc_s^2k^2\tau e^{ikc_s\tau}$. The $\tau$ cancels the $1/\tau$ in $a=-1/(H\tau)$; moreover,
$$
\mathbf k_2\cdot\mathbf k_3=\frac{k_1^2-k_2^2-k_3^2}{2}.
$$
Inserting these expressions into the <in-in formalism> and conjugating the expression inside the real part gives
$$
\boxed{
\begin{aligned}
\langle\zeta_{\mathbf k_1}(0)\zeta_{\mathbf k_2}(0)\zeta_{\mathbf k_3}(0)\rangle
={}&(2\pi)^3\delta^{(3)}(\mathbf k_1+\mathbf k_2+\mathbf k_3)
\frac{(1-c_s^2)H^4}{32\epsilon^2c_s^3(k_1k_2k_3)^3}\\
&\times\operatorname{Re}\left[
-i\int_{-\infty(1+i0^+)}^0d\tau\,e^{-iKc_s\tau}
\sum_{\rm cyc} k_1^2(k_1^2-k_2^2-k_3^2)
(1+ik_2c_s\tau)(1+ik_3c_s\tau)\right],
\end{aligned}}
$$
where $K=k_1+k_2+k_3$. Conjugating the integrand also conjugates the contour: the original positive-frequency-product integral has lower limit $-\infty(1-i0^+)$, whereas the negative exponential displayed here requires $-\infty(1+i0^+)$. This supplies the convergent interpretation of the paper's integral.
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