1. Roots of the quadratic

The equation is $x^2-x-1=0.$ Its roots are $\alpha,\beta$, so by Vieta: $\alpha+\beta=1,\qquad \alpha\beta=-1.$ Also, since $\alpha\ne\beta$ and the discriminant is $5$, we have $\alpha=\frac{1+\sqrt5}{2},\qquad \beta=\frac{1-\sqrt5}{2}.$

2. Given sequence

$a_n=p\alpha^n+q\beta^n,$ where $p,q$ are integers.

We are given $a_4=28.$ So, $p\alpha^4+q\beta^4=28.$

3. Compute $\alpha^4$ and $\beta^4$

Since $\alpha,\beta$ satisfy $x^2=x+1$, we can reduce powers: $x^2=x+1.$ Then $x^3=x(x^2)=x(x+1)=x^2+x=(x+1)+x=2x+1,$ $x^4=x(2x+1)=2x^2+x=2(x+1)+x=3x+2.$ Hence, $\alpha^4=3\alpha+2,\qquad \beta^4=3\beta+2.$

Therefore, $a_4=p(3\alpha+2)+q(3\beta+2)=3(p\alpha+q\beta)+2(p+q).$ But it is easier to substitute explicit values of $\alpha,\beta$.

4. Write $\alpha^4,\beta^4$ explicitly

Using $\alpha=\frac{1+\sqrt5}{2},\qquad \beta=\frac{1-\sqrt5}{2},$ we get \alpha^4=3\alpha+2=3\cdot\frac{1+\sqrt5}{2}+2=rac{7+3\sqrt5}{2}, \beta^4=3\beta+2=3\cdot\frac{1-\sqrt5}{2}+2=rac{7-3\sqrt5}{2}.

So $a_4=p\frac{7+3\sqrt5}{2}+q\frac{7-3\sqrt5}{2}=28.$ Multiply by $2$: $p(7+3\sqrt5)+q(7-3\sqrt5)=56.$ Expanding, $7(p+q)+3(p-q)\sqrt5=56.$

5. Use the given fact

Since $56$ is rational, comparing rational and irrational parts, $3(p-q)=0\implies p=q.$ And $7(p+q)=56\implies p+q=8.$ Since $p=q$, we get $p=q=4.$

6. Find $p+2q$

$p+2q=4+2(4)=12.$

Therefore, the correct option is $\boxed{12}.$