Answer
See a detailed answer below.
Work Step by Step
Let's assume that the liquid in the pipe is identical and noncompressible. So we can use Bernoulli’s equation of
$$ P_0+\frac{1}{2}\rho v_0^2+\rho g y_0=P_1+\frac{1}{2}\rho v_1^2+\rho g y_1$$
As we see in the figure below, the right side of the pipe extends vertically in both directions uniformly in the $y$-axis. So, $y_1=-h+h+y_0=y_0$.
So they cancel each other from Bernoulli’s formula above.
$$ P_0+\frac{1}{2}\rho v_0^2 =P_1+\frac{1}{2}\rho v_1^2 $$
Solving for $P_1$;
$$P_1 = P_0+\frac{1}{2}\rho( v_0^2 - v_1^2 ) \tag 1$$
Now we need to find $v_1$ in terms of $v_0$. We can use the equation of continuity $v_1A_1 = v_0A_0$, so
$$v_1=\dfrac{v_0A_0}{A_1}$$
where $A_1=\pi r_1^2$ where $r_1=d_1/2$, and $A_0=\pi r_0^2$ where $r_0=d_0/2$.
Thus,
$$v_1=\dfrac{v_0\color{red}{\bf\not} \pi \left[\frac{d_0}{\color{red}{\bf\not} 2}\right]^2}{\color{red}{\bf\not} \pi \left[\frac{d_1}{\color{red}{\bf\not} 2}\right]^2}$$
$$v_1=\dfrac{v_0 d_0^2}{ d_1^2}$$
Plugging into (1);
$$P_1 = P_0+\frac{1}{2}\rho\left( v_0^2 - \left[\dfrac{v_0 d_0^2}{ d_1^2}\right]^2 \right) $$
$$\boxed{P_1 = P_0+\frac{1}{2}\rho v_0^2\left(1 - \dfrac{ d_0^4}{ d_1^4} \right) } $$
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b) Here we just need to plug the given (and the known) into the boxed formula above.
$$ P_1 = (50\times 10^3)+\frac{1}{2}(1000)(10)^2\left(1 - \dfrac{ (0.168)^4}{ (0.20)^4} \right) =75106.4\;\rm Pa $$
$$P_1=\color{red}{\bf 75.1}\;\rm kPa$$
