nondimensionalization of predator-prey model

Given a set of differential equations for a predator-prey model (Lotka-Volterra): \begin{align*} \frac{dH}{dt} &= bH - sHP \\ \frac{dP}{dt} &= -dP + esHP \\ \end{align*} Where $P$ is the number of predators, $H$ the number of preys, $t$ time, $b$ the birth rate of the prey, $d$ the death rate of the predator, $s$ the searching efficiency of the predator and $e$ the efficiency in which food is turned into extra predators.

I can write this in nondimensional form using the following parameterization: \begin{align*} h &= \frac{Hes}{d} \\ p &= \frac{Ps}{b} \\ \tau &= \sqrt{bd}t \\ \rho &= \sqrt{\frac{b}{d}} \end{align*}

To become: \begin{align*} \frac{dh}{d\tau} &= \rho h (1-p) \\ \frac{dp}{d\tau} &= - \frac{1}{\rho} p (1 - h) \\ \end{align*}

My question is, how do you get this result? I understand the basics of nondimensionalization. First I find the dimensions of each variable and next I define new variables from products of variables with the same dimension. However, I think the dimensions of all variables/constants are as follows:

$H,P$ in units, $t$ in time and $b,s,d,e$ in inverse of time.

But this gets me nowhere.

Sometimes it's obvious what scales to use; for example, if there's a logistic growth term $rN(1-N/K)$, then it's natural to measure the population size $N$ in units of the carrying capacity $K$, and time in units of the reciprocal growth rate $1/r$.

In other cases, like here, it's not as obvious, but what you can always do is plug in undetermined scales (to be determined later) in the equations. Taking $$ H = c_1 h, \qquad P = c_2 p, \qquad t = c_3 \tau, $$ the system becomes $$ \begin{align*} \frac{c_1}{c_3} \frac{dh}{d\tau} &= b (c_1 h) - s (c_1 h) (c_2 p), \\ \frac{c_2}{c_3} \frac{dp}{d\tau} &= -d (c_2 p) + es (c_1 h) (c_2 p), \\ \end{align*} $$ which can be rearranged to $$ \begin{align*} \frac{dh}{d\tau} &= (c_3 b) h - (c_2 c_3 s) h p, \\ \frac{dp}{d\tau} &= -(c_3 d) p + (c_1 c_3 es) h p. \\ \end{align*} $$ Now you try to choose $c_1$, $c_2$, $c_3$ so that the coefficients become as simple as possible. Usually there is no canonical choice which gives the absolutely simplest result, but rather there are many choices which all lead to "equally simple" equations. (Sometimes the choice matters, depending on what you want to do with the equations later, but I don't think you need to worry about this here.)

There are four coefficients, but only three choices to be made, so you can't expect to make all coefficients equal to one, but you could for example choose $c_3=1/b$ and then choose $c_1$ and $c_2$ to make $c_1 c_3 es=1$ and $c_2 c_3 s=1$. This would give the system $$ \begin{align*} \frac{dh}{d\tau} &= h - h p, \\ \frac{dp}{d\tau} &= -(d/b) p + h p, \\ \end{align*} $$ where you can call the remaining coefficient $d/b$ something (say $\alpha$) to get the final form of the equations.

Now, this choice wasn't the one made by whoever gave you your equations. Instead, they opted for making the coefficients of $h$ and $-hp$ equal in the first equation, and the coefficients of $-p$ and $hp$ equal in the second equation. This requires $b=c_2 s$ and $d=c_1 es$. (Compare this to your given formulas relating $P$ and $H$ to $p$ and $h$, remembering that we have $H=c_1 h$ and $P=c_2 p$.) Finally, $c_3$ was for some reason chosen to make the overall factor $\rho$ in the first equation equal to the reciprocal of the overall factor $1/\rho$ in the second equation. Again, this is not a canonical choice but rather arbitrary.