The variation of the product is
$$
\Delta[\phi(\mathbf k)\phi(\mathbf k')]
=c(2\pi)^3\left[
\delta^{(3)}(\mathbf k)\phi(\mathbf k')
+\delta^{(3)}(\mathbf k')\phi(\mathbf k)
\right].
$$
Its expectation value vanishes because the background has $\langle\phi\rangle=0$. The <Ward identity> and part b therefore give the <shift-symmetry soft theorem for a scalar bispectrum>
$$
\boxed{\lim_{q\to0}
\frac{B_\phi(q,k,|\mathbf k+\mathbf q|)}{P_\phi(q)}=0}.
$$
Thus an exact internal shift creates no leading soft response of the two hard modes.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 312 / 2 / c / Solution

If a scalar vacuum is invariant under an exact <scalar-field shift symmetry>, the Ward identity makes the leading soft bispectrum vanish:
$$
\lim_{q\to0}\frac{B_\phi(q,k,|\mathbf k+\mathbf q|)}{P_\phi(q)}=0.
$$

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