I agree with [Brian Moths][1] that $p \rightarrow 0$ means going to small energies. 

However, I do not understand the following statement.

> as you integrate out the wavenumbers between $𝑀$ and $p$, the coupling constant $\bar{\lambda}$ gets multiplied by a large factor to give the renormalized coupling $\lambda$.

In the example given in the book, we have a positive $\beta = \frac{3\lambda^2}{16\pi^2}$. Therefore the factor is in fact smaller than 1 as we move to smaller energies ($p \rightarrow 0$).

This means that if we fix $\bar{\lambda}$ at $M$ and move to smaller energies ($p \rightarrow 0$), the effective $\lambda$ should become smaller and smaller. Now if we instead fix $\lambda$ as we move to smaller energies ($p \rightarrow 0$), we would have to make $\bar{\lambda}$ bigger and bigger to make the equation hold.

For reference, see the sentence immediately following (12.28) on Page 404.


  [1]: https://physics.stackexchange.com/users/23785/brian-moths