Work:

P2-P1 = the pressure just above the river = -sw*h, where sw = specific weight of glacier and h = height of glacier.
P2 = pressure at a depth of 500 feet = -sw*h + P1
(P1 = 1 atm = 2116 lbf/ft^2)
P2 = -(600 lbf/ft^3)(-500 ft) + 2116 lbf/ft^2
P2 = 32,116 lbf/ft^2
P2 = 223.1 psia

P4-P3 = pressure at a depth of 501 feet
P4 = -sw*h + P2
P4 = -(600 lbf/ft^3)(-1 ft) + 32,116 lbf/ft^2
P4 = 600 lbf/ft^2 + 32,116 lbf/ft^2
P4 = 32,716 lbf/ft^2
P4 = 15.5 atm
P4 = 227.3 psia

Solution

P4 = pressure at a depth of 1 foot in the river = 227.3 psia