As @DavidESpeyer [suggests](https://mathoverflow.net/questions/417698/order-of-a-rational-function-on-mathbbf-p#comment1072032_417698), I think you meant ${}+ a$ in place of ${}- a$.  As @KevinCasto [says](https://mathoverflow.net/questions/417698/order-of-a-rational-function-on-mathbbf-p#comment1072031_417698), you are then looking for the order of $\begin{bmatrix} 1 & a \\ 1 & 1 \end{bmatrix}$ as an element of $\operatorname{PGL}_2(\mathbb F_q)$, i.e. (since its eigenvalues are $1 \pm \sqrt a$), the order of $1 \pm \sqrt a$ as an element of $\mathbb F_{p^2}^\times/\mathbb F_p^\times$.  Certainly this order divides $p + 1 = \lvert\mathbb F_{p^2}^\times/\mathbb F_p^\times\rvert$; but, as @DavidESpeyer also [suggests](https://mathoverflow.net/questions/417698/order-of-a-rational-function-on-mathbbf-p#comment1072032_417698), it need not equal $p + 1$.  Indeed, if $p = 5$ and $a = 2$, then the element has order $3$, since $\begin{pmatrix} 1 & a \\ 1 & 1 \end{pmatrix}^3 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$.