b
Apply the equation of continuity.
$A _{1} v _{1}= A _{2} v _{2}$
$5^{2} \times 4=2^{2} \times v _{2}$
$v _{2}=25$
Apply the energy equation at both the ends.
$P _{1}+\frac{1}{2}\rho v _{1}^{2}= P _{2}+\frac{1}{2}\rho v _{2}^{2}$
$P_{1}-P_{2}=\frac{1}{2}\rho \left(v_{2}^{2}-v_{1}^{2}\right)$
$=\frac{1}{2} \times 10^{3}\left(25^{2}-4^{2}\right)$
$=304500 Pa$