Define $b_k=\frac{a_k}{P(a_k)}$. It is easy to see that for any prime $p \mid a_k$, we must have $p \mid a_{k+1}$ $\implies$ $P(a_k) \mid P(a_{k+1})$. Hence we get $$b_{k+1}=\frac{P(a_k)}{P(a_{k+1})}(b_k+1) \leq b_k+1$$So we are done by discrete continuity if $b_n \leq 2015$ and $b_m \geq 2015$ for some $n \leq m$. So we have two cases left: Case I: $b_n >2015$ for all $n$. By suitably shifting the sequence, we can assume that $b_0= \min \{b_0,b_1, \ldots \}$. Let $m$ be the smallest non-negative integer such that $$\frac{P(a_{m+1})}{P(a_m)}= t \geq 2$$Hence we have $P(a_0)=P(a_1)=\cdots=P(a_m)$. Call this common value $R$. Note that $\gcd(R,t)=1$. The original equation gives us $b_i=b_0+i$ for $0 \leq i \leq m$. Also by our assumption $$b_0 \leq b_{m+1} = \frac{b_m+1}{t}$$$\implies$ $m \geq (t-1)b_0-1$. So $b_0,b_1, \ldots b_m$ form a set of at least $(t-1)b_0$ consecutive integers, and hence every prime not exceeding $(t-1)b_0$ divides at least one of those terms. But $P(b_i) \mid P(a_i)=R$ for all $i \leq m$ $\implies$ $R$ is divisible by every prime not exceeding $(t-1)b_0$ $\implies$ $t>(t-1)b_0$, which is absurd because $t \geq 2$ and $b_0 > 2015$. $\blacksquare$ Case II: $b_n < 2015$ for all sufficiently large $n$. By suitably shifting the sequence, we can assume that $b_n <2015$ for all $n$. Call a prime $p$ new if $p \nmid a_0$ but $p \mid a_i$ for some $i \geq 1$. Since $a_0, a_1, \dots$ is unbounded and $b_0,b_1, \dots$ is bounded, the sequence $P(a_0),P(a_1), \dots$ must be unbounded. Therefore there must be infinitely many new primes. Choose any new prime $p>2016$, and let $m \geq 0$ such that $p \nmid a_m$ but $p \mid a_{m+1}$ $\implies p \mid P(a_m)(b_m+1)$ $\implies$ $p \mid b_m+1$ $\implies$ $p \leq b_m+1 < 2016$. Contradiction! $\blacksquare$

Umm.. is it safe to assume that the number $2015$, can be replaced with any positive integer $r$?