Answer
$P_{1}=17.75psia$
Work Step by Step
Starting at point $1$ in the natural gas pipeline and assuming that the variation on pressure cause by the air is negligible:
$P_{1}-\rho_{Hg}*g*h_{Hg}+\rho_{oil}*g*h_{oil}-\rho_{water}*g*h_{water}=P_{atm}$
$P_{1}=P_{atm}+g*(\rho_{Hg}*g*h_{Hg}-\rho_{oil}*g*h_{oil}+\rho_{water}*g*h_{water})$
$\rho_{water}=62.4\frac{lbm}{ft^3}$
$\rho_{Hg}=13.6*62.4\frac{lbm}{ft^3}=848.64\frac{lbm}{ft^3}$
$\rho_{oil}=0.69*62.4\frac{lbm}{ft^3}=43.06\frac{lbm}{ft^3}$
$h_{water}=(25in+2in)*(\frac{1ft}{12in})=2.25ft$
$h_{Hg}=(6in)*(\frac{1ft}{12in})=0.5ft$
$h_{oil}=(25in-10in)*(\frac{1ft}{12in})=1.25ft$
Substituting:
$P_{1}=14.2psia+(32.2\frac{ft}{s^2})*[(848.64\frac{lbm}{ft^3})*(0.5ft)-(43.06\frac{lbm}{ft^3})*(1.25ft)+(62.4\frac{lbm}{ft^3})*(2.25ft)]*(\frac{1lbf}{32.2\frac{lbm*ft}{s^2}})*(\frac{1ft^2}{144in^2})$
$P_{1}=17.75psia$
