The second term on the rhs disappears since this is just the mean and employing the definition of \( \sigma^2 \) we have
$$ \int_{-\infty}^{\infty}dxp(x)e^{\left(iq(\mu-x)/m\right)}= 1-\frac{q^2\sigma^2}{2m^2}+\dots, $$resulting in
$$ \left[\int_{-\infty}^{\infty}dxp(x)\exp{\left(iq(\mu-x)/m\right)}\right]^m\approx \left[1-\frac{q^2\sigma^2}{2m^2}+\dots \right]^m, $$and in the limit \( m\rightarrow \infty \) we obtain
$$ \tilde{p}(z)=\frac{1}{\sqrt{2\pi}(\sigma/\sqrt{m})} \exp{\left(-\frac{(z-\mu)^2}{2(\sigma/\sqrt{m})^2}\right)}, $$which is the normal distribution with variance \( \sigma^2_m=\sigma^2/m \), where \( \sigma \) is the variance of the PDF \( p(x) \) and \( \mu \) is also the mean of the PDF \( p(x) \).