Clearly, if $p=2,5,19$, then we can't find such $a_n$. Hence, $p\in \{3,7,11,13,17\}$. If $p=11$, then $a_n=2 \cdot 10^{n+1}+19\equiv 0\pmod{11}\Rightarrow 10^{n+1}\equiv 7\pmod{11}$. But, $10^{n+1}\equiv (-1)^{n+1}\equiv \{1,-1\}\pmod{11}$. Contradiction. Hence, $p\in \{3,7,13,17\}$. For $p=3$, we have $2\cdot 10^2+19=219\equiv 0\pmod{3}$. $(n=1)$ For $p=7$, we have $2\cdot 10^6+19\equiv 2\cdot 1+19\equiv 21\equiv 0\pmod{7}$. $(n=5)$ For $p=13$, we have $2\cdot 10^{13}+19\equiv 2\cdot 10+19\equiv 39\equiv 0\pmod{13}$. $(n=12)$ For $p=17$, we have $2\cdot 10^8+19\equiv 2\cdot 100^4+19\equiv 2\cdot 15^4+19\equiv 2\cdot 225^2+19\equiv 2\cdot 4^2+19\equiv 51\equiv 0\pmod{17}$. $(n=7)$ Thus, our answer is: there exists some $n \ge 1$ such that $p$ divides $a_n$ $\Leftrightarrow p\in \{3,7,13,17\}$.