If SAT is in PCP, for some constant q, then P = NP I have seen this statement before, but I haven't really seen a proof of it:

If $SAT\in PCP_{1,2^{−q}}[\log(n),q]$, for some constant $q$, then $P = NP$.

Now, if $SAT\in PCP_{1,2^{−q}}[\log(n),q]$, then SAT reduces to $GAP-q-CSP[1,2^{-q}]$. So to show $P = NP$, how do we show then that $GAP-q-CSP[1,2^{-q}]$ is in $P$?
Given a CSP where all constraints have arity at most $q$ we want to distinguish between the case where everything is satisfiable and the case where at most $1/2^q$ fraction of the constraints are satisfiable, in polynomial time. Here is how this can be done.

First, all predicates used in the CSP must have at least one satisfying assignment (otherwise we know that the instance is not perfectly satisfiable and we are done). If a used predicate has two or more satisfying assignments, then, if we take a random assignment for all variables some constraint is satisfied with probability at least $2/2^q&gt;1/2^q$. Since all other constraints are satisfied with probability at least $1/2^q$, by linearity of expectations there exists an assignment that satisfies strictly more than $1/2^q$ fraction of the constraints and we are done in this case.

Finally, suppose that all predicates used have exactly one satisfying assignment. Then the CSP is an instance of DNF-SAT, which can be decided in polynomial time (we only need to check if any two constraints have a conflict).