Answer
$P_{p}=0.732kW$
Work Step by Step
Here we are working with potential energy:
The potential work is: $W_{p}=mgh$
And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=\rho Qgh$
The potential power needed is:
$P_{p}=1000\frac{kg}{m^3}*(320\frac{L}{min}*\frac{1m^3}{1000L}*\frac{1min}{60s})*9.81\frac{m}{s^2}*14m$
$P_{p}=0.732kW$