First we solve the problem for points on a straight line instead of a polygon. Suppose there are $n$ points on a straight line. Let $f(n,k)$ denote the number of ways to color exactly $k$ points red such that no consecutive vertices are both red. Now, we find a recursion for $f(n,k)$. If the $nth$ point is red, then the $(n-1)$th point cannot be red. The rest can be colored $f(n-2, k-1)$ ways. If the $nth$ point is not red, then the remaining can be colored $f(n-1, k)$ ways. Therefore; $$ f(n,k)= f(n-1,k)+f(n-2,k-1)$$ It is easy to prove by induction that $f(n,m)=\binom{m-n+1}{n}$ Returning to the original problem, fix a vertice P. If it red, the two adjacent ones cannot be red, The rest can be colored $f(n-3, k-1)$ ways. If P is not red, then the rest can be colored $f(n-1,k)$ ways. Therefore the required answer is- $$f(n-3, k-1) + f(n-1,k) =\binom{n-k-1}{k-1}+\binom{n-k}{k}$$