If p(n) is a polynomial in n, then is 1/p(n) polynomially bounded?
In order to estimate the lower bound on $k$, you use the estimate $1/(1+\epsilon) = 1-\epsilon+\Theta(\epsilon^2)$, which is true as $\epsilon\to0$ (which is certainly your case). Using this, we can analyze the denominator:
$$
\begin{align*}
\bar{p}-1+\frac{1}{4\bar{p}} &amp;= \frac{1}{p(n)}-\frac{1}{2}+\frac{1}{2}\frac{1}{1+2/p(n)} \\ &amp;=
\frac{1}{p(n)}-\frac{1}{2}+\frac{1}{2}\left(1 - \frac{2}{p(n)} + \Theta\left(\frac{1}{p(n)^2}\right)\right) \\ &amp;=
\Theta\left(\frac{1}{p(n)^2}\right).
\end{align*}
$$
Therefore the lower bound on $k$ is of order of magnitude $$\Theta(np(n)^2).$$