Please Excuse My Dear Ant Brilli

Slope of the line segment $\overline {AB}$ is $\dfrac{\dfrac{1}{1^p}+\dfrac{1}{3^p}+\dfrac{1}{5^p}+...}{\dfrac{1}{2^p}+\dfrac{1}{4^p}+\dfrac{1}{6^p}+...}=2^p-1=\dfrac{1}{16}\times {496}$ . $496$ is a perfect number. So $2^p-1=31$ or $2^p=32=2^5$ . Hence $p=\boxed 5$ .

(If $p$ and $2^p-1$ be primes, then $2^{p-1}(2^p-1)$ is a perfect number. So, for prime $p, 2^p-1$ is $(\dfrac{1}{2^{p-1}})^{th}$ of a perfect number. Hence $2^{p-1}=16=2^4$ , and from here, $p=5$ , a prime number.)

Assume an unorthodox notation, where $n' = \frac 1{n^p}$ , e.g. $2' = \frac 1{2^p}$ .
Let $\displaystyle \text{EV}=\sum_{n=0}^\infty \frac 1{(2n)^p} = 2'+4'+6'+8'+\cdots$
and $\displaystyle\text{OD}=\sum_{n=0}^\infty \frac 1{(2n-1)^p} =1' +3'+5'+7'+\cdots$ .

Note that

$\begin{aligned}\\ \frac {\text{EV}}{\text{OD}} &=\frac {2'+4'+6'+8'+\dots} {1'+3'+5'+7+\cdots} \\\\ &=\frac {2' (1'+2'+3'+4'+\dots) } {1'+3'+5'+7+\cdots} \\\\ &=\frac {2' \big[\text{OD}+\text{EV} \big] } {\text{OD}} \\\\ &=2'\left(1+\frac {\text{EV}}{\text{OD}} \right) \\\\ (1-2')\frac {\text{EV}}{\text{OD} } &= 2'\\\\ \text {Slope of }AB &=\frac {\text{OD}}{\text{EV} }\\\\ &=\frac {1-2'}{2'}\\\\ &=\frac 1{2'}-1\\\\ &=2^p-1 \end{aligned}$

It is given that
$\begin{aligned} \text{Slope of} AB &= \frac 1{16} N &&(N = \text{perfect number})\\ \therefore 2^p-1 &=\frac 1{16} \cdot \overbrace{2^{q-1} (2^q-1)}^N &&(q \text{ prime}) \end{aligned}$

Therefore, $2^{q-1}=16$ and $p=q$ i.e. $p(=q)=\color{#D61F06}5$

The ant goes up $\Delta y = \displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^p}$ and right $\Delta x = \displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n)^p}$ .

Let $k = \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$ . Then $\Delta x = \displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n)^p} = \frac{1}{2^p}\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p} = \frac{1}{2^p}k$ , and $\Delta y = k - \Delta x = k - \frac{1}{2^p}k = \displaystyle \frac{2^p - 1}{2^p}k$ .

The slope of $AB$ is therefore $m = \frac{\Delta y}{\Delta x} = \displaystyle \frac{\frac{2^p - 1}{2^p}k}{\frac{1}{2^p}k} = 2^p - 1$ .

Since the slope $m$ is equal to $\frac{1}{16}$ of a perfect number $P$ , then $P = 16m = 16(2^p - 1)$ .

Since all even perfect numbers are in the form of $2^{p - 1}(2^p - 1)$ and $16$ is its only even factor, $16 = 2^{p - 1}$ , and this solves to $p = \boxed{5}$ .