I feel obliged to flesh out my comments (and to modify my wrong answer, thanks to GH from MO).
Write $n=ab$, and let $H_{k}$ the $k$-th harmonic number, $\tau(n)$ the number of positive divisors of $n$. LHS is greater than the related sum $\sum_{a=1}^{p-1}\sum_{b=1}^{a}\dfrac{b}{a(ab)_{p}}$, that we'll denote by $S$.
Provided the limit as $\tau(n)$ tends to $\infty$ of $f(n):=\frac{2}{\tau(n)}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{d^{2}}{n}$ exists and equals a positive constant $M$, we have:
$S=\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{d^2}{n(n)_{p}}\sim\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{M}{(n)_{p}}\sim\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}M\dfrac{H_{p-1}}{p-1}\sim M\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{\log(p-1)}{p-1}$.
The idea is to consider all divisors of $n$ less than its square root, and to replace $\dfrac{1}{n_p}$ by its average value, which is $\dfrac{H_{p-1}}{p-1}\sim\dfrac{\log(p-1)}{p-1}$ (in my comment I erroneously took the reciprocal of the average value and not the average value of the reciprocals, hence the missing $\log(p-1)$).
We thus obtain $S\sim\dfrac{\log(p-1)}{p-1}\sum_{n=1}^{(p-1)^2}\dfrac{\tau(n)}{2}$. As $D(x):=\sum_{n=1}^{x}\tau(n)$ is provably asymptotic to $x(\log x+K)$ where $K$ is a positive constant (see Dirichlet divisor problem on Wikipedia), we end up with:
$S\sim\frac{M}{2}\frac{\log(p-1)}{p-1}(p-1)^2(\log (p-1)^2+K)\sim Mp\log^{2} p$ which provides a lower bound for the sum of the OP of the desired order of magnitude.