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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 $S\sim M\dfrac{\log(p-1)}{p-1}\sum_{n=1}^{(p-1)^2}\dfrac{\tau(n)}{2}$. 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.

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 $S\sim M\dfrac{\log(p-1)}{p-1}\sum_{n=1}^{(p-1)^2}\dfrac{\tau(n)}{2}$. 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.

Edit: the following link: http://www.les-mathematiques.net/phorum/read.php?5,1967230,1967504#msg-1967504 shows that if $M$ exists, it equals $\frac{1}{2}$.

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.

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 $S\sim M\dfrac{\log(p-1)}{p-1}\sum_{n=1}^{(p-1)^2}\dfrac{\tau(n)}{2}$. 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.

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 $\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{d^2}{n(n)_{p}}\leqslant\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{1}{(n)_{p}}\lesssim\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{H_{p-1}}{p-1}\lesssim\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{\log(p-1)}{p-1}$.

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 $LHS\lesssim\dfrac{\log(p-1)}{p-1}\sum_{n=1}^{(p-1)^2}\dfrac{\tau(n)}{2}$. 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:

$LHS\lesssim\frac{1}{2}\frac{\log(p-1)}{p-1}(p-1)^2(\log (p-1)^2+K)\lesssim p\log^{2} p$ which is the desired order of magnitude.

Edit: the desired asymptotics would follow from $\frac{2}{\tau(n)}\sum_{d\mid n,d\leq\sqrt{n}}\frac{d^2}{n}\sim\frac{1}{2}$ for large values of $\tau(n)$.

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.