What do the field operators $\psi$ and $\pi$ produce when they act on vacuum $|0>$ state? Here,
$$\psi(\vec{x}) = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_p}}\left(a_p e^{i\vec{p}\cdot \vec x} + a_p^\dagger e^{-i\vec p \cdot \vec x}\right) $$
and
$$\pi(\vec{x}) = -i\int \frac{d^3p}{(2\pi)^{3}}\sqrt{\frac{E_p}{2}}\left(a_p e^{i\vec p \cdot \vec x} - a_p^\dagger e^{-i\vec p \cdot \vec x}\right)$$
Your definitions/conventions trivially yield $$\psi(\vec{x})|0\rangle = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{ 2E_p } e^{-i\vec p \cdot \vec x} |\vec p\rangle $$ and $$\pi(\vec{x})|0\rangle = \frac{i}{2}\int \frac{d^3p}{(2\pi)^{3}} e^{-i\vec p \cdot \vec x} | \vec p\rangle .$$ What is it you are seeking? These are wavepackets of one-particle states localized at x, within basically a Compton wavelength 1/m, as you might know. You may appreciate this near-duplicate answer. P&S (2.35) define $$ |\vec p\rangle\equiv \sqrt{2E_p} ~a_p^\dagger|0\rangle,\implies \\ \langle \vec p| \psi (\vec x)|0\rangle = e^{-i \vec p \cdot \vec x}. $$
