, 2016

Let $c$ denote the other root of $Q$. Since $P$ and $Q$ have rational coefficients the Vieta's formulas imply
$$\begin{aligned} a+b &= p_1 \in \mathbb{Q} \quad \text{and} \quad ab = p_2 \in \mathbb{Q}, \\ (a+2016)+c &= q_1 \in \mathbb{Q} \quad \text{and} \quad (a+2016)c = q_2 \in \mathbb{Q}. \end{aligned}$$
Expressing $b = p_1 - a$ and $c = q_1 - a - 2016$ from the equations on the left and inserting them into the equations on the right we get
$$\begin{aligned} a(p_1 - a) &= p_2, \\ (a + 2016)(q_1 - a - 2016) &= q_2. \end{aligned}$$
Subtracting one equation from the other we get
$$\begin{aligned} a(p_1 - a) - (a + 2016)(q_1 - a - 2016) &= p_2 - q_2 \\ \Rightarrow a(p_1 - q_1 + 4032) &= p_2 - q_2 + 2016q_1 - 2016^2 \in \mathbb{Q}. \end{aligned}$$
If $p_1 - q_1 + 4032 \neq 0$, we can write
$$a = \frac{p_2 - q_2 + 2016q_1 - 2016^2}{p_1 - q_1 + 4032} \in \mathbb{Q}.$$
It would follow that $a$ is rational, which is not the case. So, $p_1 - q_1 + 4032 = 0$. From here
$$(a+b) - (a+2016+c) + 4032 = 0 \Rightarrow c = b+2016.$$

Hence, the only candidate for the other root of $Q$ is the number $b + 2016$. We have to check that there indeed exists a polynomial $Q$ with rational coefficients such that $a + 2016$ and $b + 2016$ are its roots. An example of such a polynomial would be
$$Q(x) = (x - a - 2016)(x - b - 2016) = x^2 - (a + b)x + ab + 2016(a + b) + 2016^2,$$
since $a + b \in \mathbb{Q}$ and $ab + 2016(a + b) + 2016^2 \in \mathbb{Q}$. The only possible value for the other root of the polynomial $Q$ is therefore $b + 2016$.