Not exactly an answer either but I feel obliged to flesh out my comments.
Write $n=ab$, and let $H_{k}$ the $k$-th harmonic number, $\tau(n)$ the number of positive divisors of $n$. LHS becomes $\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{d^2}{n(n)_{p}}\leqslant\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{1}{(n)_{p}}\ll\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{H_{p-1}}{p-1}\ll\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{\log(p-1)}{p-1}$.
The idea is to upper bound the inner sum by taking all divisors of $n$ less than its square root and replace this inequality by an equality, 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 $LHS\ll\dfrac{\log(p-1)}{p-1}\sum_{n=1}^{(p-1)^2}\tau(n)$. 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:
$LHS\ll\frac{\log(p-1)}{p-1}(p-1)^2(\log (p-1)^2+K)\ll p\log^{2} p$ which is the desired order of magnitude.