Power is work per second, and the work a pump does is lifting weight. So the hydraulic power follows straight from the two numbers you already have: $P = \rho g Q H$ — read aloud P equals rho g Q H. $P$ is the hydraulic power in watts, the power actually delivered to the liquid; $\rho$ is the liquid's density in kg/m³ (1000 for water); $g$ is 9.81 m/s²; $Q$ is the flow in cubic metres per second; and $H$ is the total head in metres — the number the last lesson assembled.

Read $\rho g Q$ as one idea and the formula stops being algebra: it is the weight of liquid crossing the flange every second, in newtons per second. Multiply by the height it is being lifted and you have joules per second, which is watts. Nothing else is going on.

The unit trap is $Q$. Plant rooms talk in litres per second and the formula wants cubic metres per second — a factor of a thousand, and it lands in the answer as a thousandfold error that looks perfectly plausible on the page. Fold it in once and carry the shortcut: for water, $P\ (\mathrm{kW}) = \dfrac{Q\ (\mathrm{L/s}) \times H\ (\mathrm{m})}{102}$, because 9.81 over a thousand is one over 102. That 102 is the metric cousin of the imperial trade's $\dfrac{Q \, H \, SG}{3960}$; both are ρg with the unit conversions baked in, and neither is a law of nature.

One more piece of housekeeping. Specific gravity $SG$ is a bare ratio — the liquid's density over water's — so pumping 1.2 SG brine at the same flow and head costs 20 % more power. In SI you simply put the real $\rho$ in and the arithmetic does it for you.