1. Let the four consecutive coefficients in the expansion of $(1+x)^n$ be $\binom{n}{r},\ \binom{n}{r+1},\ \binom{n}{r+2},\ \binom{n}{r+3}.$

   According to the question, $\binom{n}{r}=2-p,\quad \binom{n}{r+1}=p,\quad \binom{n}{r+2}=2-\alpha,\quad \binom{n}{r+3}=\alpha.$

2. Since coefficients in the expansion of $(1+x)^n$ are positive integers, all these numbers must be positive integers.

   Also, $\binom{n}{r}+\binom{n}{r+1}=(2-p)+p=2,$ and $\binom{n}{r+2}+\binom{n}{r+3}=(2-\alpha)+\alpha=2.$

3. Now, among binomial coefficients, the only positive integers whose sum is $2$ are $1$ and $1$.

   Hence, $2-p=1,\quad p=1$ and $2-\alpha=1,\quad \alpha=1.$

4. Therefore, $p=1,\quad \alpha=1.$

   Now compute: $p^2-\alpha^2+6\alpha+2p=1^2-1^2+6(1)+2(1).$

   $=1-1+6+2=8.$

5. Hence the required value is $\boxed{8}.$

So the correct option is A.