I can prove that $\sum^{P}_{k=1} \lfloor\frac {ka}{p}\rfloor = \mu(a,p)$ mod$(2)$ where $p$ is an odd prime, $P = \frac {p-1}{2}$, $a$ is an integer not divisible by $p$, and $\mu(a,p)$ is the quantity defined as the number of integers in the list $a, 2a, 3a, ..., Pa$ that become negative when the integers in the list are reduced modulo $p$ into the interval from $-P$ to $P$.
How do I show that, when $a$ is even instead of odd, and still not divisible by $p$,
$\sum^{P}_{k=1} \lfloor\frac {ka}{p}\rfloor = \frac{p^{2} - 1}{8} + \mu(a,p)$ mod$(2)$
is true?
Using the last result I proved in your previous question, namely that for any odd prime $p$ and any natural number $b$ not divisible by $p$ we have:
$$\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{2bk}{p}}\right\rfloor\equiv (b-1)\frac{p^2-1}{8}+\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{bk}{p}}\right\rfloor\text{ mod } 2$$
We can then set $b=a$ an even integer so that $(b-1)\equiv (a-1)\equiv 1 \text{ mod } 2$ which gives us:
$$\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{2ak}{p}}\right\rfloor\equiv \frac{p^2-1}{8}+\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{ak}{p}}\right\rfloor\text{ mod } 2$$