I think you got the sign of the exponential in Schwartz's (28.8) wrong here, which I'll reverse, but I am cavalier with signs and factors, as I assume you just want to see the principle (the forest, not the trees). 

The key point is that the SSB current is basically *always* of the form $j_\mu\propto F \partial_\mu \pi (x) +...$ where the omitted terms are of higher older in the fields, and of course group indices have been omitted as well. This is codified in the appellation "Nambu-Goldstone nonlinear realization of the symmetry". This is the only way to have the v.e.v. of the transform of this Goldstone field not vanish, while the v.e.v.s of the fields themselves vanish--after shift redefinitions. (See [this answer](https://physics.stackexchange.com/a/369218/66086) .)

The Fourier transform of the Goldstone field π is 
$$
\pi(\vec{\bf{p}}) = \frac{-2i}{F}\int d^3x ~e^{i\,\vec{\bf{p}}\cdot\vec{\bf{x}}}j^0(x),
$$  
with 
$$
[H,\pi(\vec{\bf{p}})] =  E(\vec{\bf{p}}) ~  \pi(\vec{\bf{p}})  .
$$

As a consequence, decompressing your text's one liner,
$$H|\pi(\vec{\bf{p}})\rangle= (   [H,\pi(\vec{\bf{p}})] + \pi(\vec{\bf{p}}) H ) ~|\Omega\rangle =
( E(\vec{\bf{p}}) + E_0 )\pi(\vec{\bf{p}})|\Omega\rangle= ( E(\vec{\bf{p}}) + E_0 )|\pi(\vec{\bf{p}})\rangle .$$