Notice that any number after the comma is of the form $x\cdot 10^{-y}$ for some $x,y\in \mathbb{Z}^+$, thus they are rational. So the remaining number is rational as it is the difference of rational numbers. Suppose wlog $y_1<y_2<…<y_{2012}$ Then $\displaystyle \frac{a}{b}=\frac{1}{p}-\frac{x_1}{10^{y_1}}-...-\frac{x_{2012}}{10^{y_{2012}}}=\frac{10^{y_{2012}}-p\left(x_1\cdot 10^{y_{2012}-y_1}+...+x_{2012}\right)}{p\cdot 10^{y_{2012}}}$ $p>5$ $\Rightarrow$ $p$ doesn't divide $10^{y_{2012}}$ thus $p$ can't be simplified. Hence $b$ is a multiple of $p$