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) .