From the first equation $x^2 + px + 12 = 0$ one root is given as $4$.

Substitute $x=4$ into the equation: $4^2 + 4p + 12 = 0$ $16 + 4p + 12 = 0$ $28 + 4p = 0$ $4p = -28$ $p = -7$

Now consider the second equation: $x^2 + px + q = 0$ Since $p=-7$, it becomes $x^2 - 7x + q = 0$

This equation has equal roots, so its discriminant must be zero: $b^2 - 4ac = 0$ $(-7)^2 - 4(1)(q) = 0$ $49 - 4q = 0$ $4q = 49$ $q = \frac{49}{4}$

Therefore, the correct option is $\boxed{\frac{49}{4}}$ which is option D.