$Q_A = Q_B = 0.176 \, \text{ m}^3\text{/s}$
Velocity heads
$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$
$\dfrac{{v_A}^2}{2g} = \dfrac{8(0.176^2)}{\pi^2(9.81)(0.15^4)} = 5.0557 \, \text{ m}$
$\dfrac{{v_B}^2}{2g} = \dfrac{8(0.176^2)}{\pi^2(9.81)(0.45^4)} = 0.0624 \, \text{ m}$
Neglecting head loss
$E_A = E_B$
$\dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma} + z_A = \dfrac{{v_B}^2}{2g} + \dfrac{p_B}{\gamma} + z_B$
$5.0557 + \dfrac{p_A}{\gamma} + 0 = 0.0624 + \dfrac{p_B}{\gamma} + 4.5$
$\dfrac{p_B}{\gamma} - \dfrac{p_A}{\gamma} = 0.4933 \, \text{ m}$
$p_B - p_A = 0.4933\gamma$
$p_B - p_A = 0.4933(9.81)$
$p_B - p_A = 4.839 \, \text{ kPa}$ answer
