Put $k = 2016$ and $a_{n} = \max \{p_{n}, p_{n+1}\}$ for each $n \geqslant 1$.
We shall show that $a_{n+1} \leqslant a_{n} + k + 2$. In fact, since $p_{n+1} \leqslant a_{n}$, it remains to show that $p_{n+2} \leqslant a_{n} + k + 2$.
If $p_{n+2} = 2$ then the inequality is trivially true. It suffices to consider the case $p_{n+2} > 2$.
If either $p_{n} = 2$ or $p_{n+1} = 2$ then $p_{n+2} \leqslant p_{n} + p_{n+1} + k = a_{n} + k + 2$. Otherwise, both primes $p_{n}$ and $p_{n+1}$ are odd, in particular $p_{n} + p_{n+1} + k$ is even. But $p_{n+2} > 2$, we get
$$p_{n+2} \leqslant \frac{p_{n} + p_{n+1} + k}{2} < a_{n} + k + 2 .$$
It is clear that there is a positive integer $s$ such that $a_{1} \leqslant s(k+3)! + 1$. We show, by induction that $a_{n} \leqslant s(k+3)! + 1$. Indeed, suppose this is true for $n$, then
$$a_{n+1} \leqslant a_{n} + k + 2 \leqslant s(k+3)! + k + 3 .$$
If $a_{n+1} > s(k+3)! + 1$, then $1 < m \leqslant k + 3$ with $m = a_{n+1} - q(k+3)!$. Hence, $m$ must be a divisor of $(k+3)!$, that is $m$ is a proper divisor of $s(k+3)! + m = a_{n+1}$. However, this cannot happen since $a_{n+1}$ is a prime.
Thus, $a_{n} \leqslant s(k+3)! + 1$, and this completes the solution. $\square$