Missing details: figure of the problem is attached.
We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

where:
is the pressure difference between the two ends
is viscosity of the fluid
L is the length of the pipe
is the volumetric flow rate, with
being the section of the tube and
the velocity of the fluid
r is the radius of the pipe.
We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:
is the dynamic water viscosity at 
L=4.0 cm=0.04 m

and r=1 mm=0.001 m
Using these data in the formula, we get:

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.