Spontaneously broken linear sigma model in Peskin & Schroeder: where is the miracle? P&S spend almost 12 pages discussing the renormalisability of the spontaneously broken linear sigma model and give a detailed calculation of the cancellation of divergences at one-loop level and call this a miracle.
Now I think the only thing they have done is shifted one field $\phi_N (x)\longrightarrow v+\sigma (x)$ where $v$ is a constant. If looked at from the pov of $\phi^4$ theory in four dimension we know the theory is renormalisable, so why is this not obvious for the theory $(\phi_1, \cdots , \phi_{N-1}, v+ \sigma$)? I.e. Why do they claim this is a miracle?
I understand that the symmetry of the ground state is broken as one specific direction is chosen, but why should this impact the renormalisability of the theory? I.e. Why do they claim this is a miracle?
 A: I believe they are making a pedagogical point here that for you may be intuition/obvious. 
The point of this section is that there may be more distinct terms in a Lagrangian than there are free parameters, and these distinct terms will give rise to different divergent diagrams, but if those distinct terms are related by a symmetry then so will their divergent diagrams. Therefore, you will need less renormalizing counterterms than you may expect. The authors hint at this in the last paragraph of their introduction to chapter 11.
For example, the linear sigma model Lagrangian has only $3$ parameters ($\mu^2$, $\lambda$, and an assumed common field strength) but technically $N+N+\frac{N(N+1)}{2}$ distinct terms. In such a case one might naively wonder "to renormalize this theory, do I need $3$ counterterms, or $N+N+\frac{N(N+1)}{2}$ counterterms?" Of course, there is an $O(N)$ symmetry relating the terms which brings the number of distinct terms down to $3$, so at this point it's nice but not too surprising to know that we only need $3$ counterterms to completely renormalize the theory. 
But then the situation gets hairier: What if that symmetry of the original Lagrangian is not preserved by the (true) vacuum? In that case, in order to do standard perturbation theory we will need to expand our fields/Lagrangian about the vacuum state, hiding the symmetry of the original Lagrangian. Upon doing a perturbative expansion in our new fields (to calculate, say, 1PI diagrams), we will now get different diagrams with (potentially) different divergences. This is shown to occur in the linear sigma model. Now that we no longer have the explicit symmetry of the original Lagrangian, will we need more renormalizing counterterms? The authors show that, to one-loop order in the linear sigma model, only $3$ counterterms are needed, no more.
Another point of that chapter which Dr. Zachos pointed out in his comments is the notion of anomalous spontaneous symmetry breaking, which actually (I believe) is not relevant for the "miracle" of needing only 3 counterterms, but which is nonetheless very interesting and important. The linear sigma model with $\mu^2=0$ does not have any anomalous $O(N)$ symmetry breaking, as the authors argue (the final argument for this is given in section 13.2), but there do exist other QFT's which exhibit anomalous symmetry breaking - e.g. the Coleman-Weinberg model, which you can explore in the "final project" of Part 2 of Peskin & Schroeder.
A: Renormalization means splitting parameters of theory (field strength, charge) into physical- and counter- terms. Number and type of this parameters is constrained by the symmetries of original Lagrangian.
In broken symmetry it is no longer obvious what form original Lagrangian had, even less obvious is that one would have to introduce finite number of counterterms to renormalize it since there are no obvious symmetries to limit them.
