Let's say we have something like this:
$$Pr(|\frac{Y}{n}-p|< e) \geq 1-\frac{p(1-p)}{e^2n}$$
Let's say we know that $p(1-p) \leq \frac{1}{4}$
Now if I rearrange all the terms in the first equation I get:
$$p(1-p) \geq e^2n(1-Pr(|\frac{Y}{n}-p|< e))$$
Since this shows that $p(1-p)$ is great than RHS I can safely replace p(1-p) with something that is even greater.
The question is, is there another way to know that I can immediately replace the p(1-p) with 1/4 or is the only way to rearrange all the terms to verify?
The right hand side of your original equation is just a number, depending on $p$ and $n$. By considering the quadratic $f(p) = p(1-p)$, and maximising it, we see that $f(p) \le \tfrac14$ for all $p$. Hence we can say $$ 1 - \frac{p(1-p)}{e^2n} \ge 1 - \frac{1/4}{e^2n}, $$ noting that we're looking at $-p(1-p)$, and so $f(p) \le \tfrac14$ for all $p$ becomes $-f(p) \ge -\tfrac14$ for all $p$. But we know that $P(|Y/n - p|<e)$ is at least the term in the right. Hence we have $$ P(|Y/n - p|<e) \ge 1 - \frac{p(1-p)}{e^2n} \ge 1 - \frac{1/4}{e^2n}. $$