Water is flowing through a long horizontal tube. Let P_A and P_B be the pressure at two points A and B of the tube

By Bernoulli's theorem, 

${\mathrm{P}}_{\mathrm{A}}+{\mathrm{h}}_{\mathrm{A}}\mathrm{\rho g}+\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_{\mathrm{A}}}^2={\mathrm{P}}_{\mathrm{B}}+{\mathrm{h}}_{\mathrm{B}}\mathrm{\rho g}+\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_{\mathrm{B}}}^2$

Since the pipe is horizontal, $h_A\rho g$ and $h_B\rho g$ are both zero.

as a result,

${\mathrm{P}}_{\mathrm{A}}+\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_{\mathrm{A}}}^2={\mathrm{P}}_{\mathrm{B}}+\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_{\mathrm{B}}}^2$

${\mathrm{P}}_{\mathrm{A}}-{\mathrm{P}}_{\mathrm{B}}=\frac{1}{2}\mathrm{\rho }({{\mathrm{v}}_{\mathrm{B}}}^2-{{\mathrm{v}}_{\mathrm{A}}}^2)$

Now, ${\mathrm{P}}_1 \mathrm{and} {\mathrm{P}}_2$ are equivalent if and only if velocities $v_1 and v_2$ are comparable.

Now that the cross sectional area is the same at both sites, speeds ${\mathrm{v}}_1 \mathrm{and} {\mathrm{v}}_2$ are equal.

$\Rightarrow {\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_{\mathrm{B}}$when cross-sectional area at A and B is same