If one of the resistors becomes open circuited, what is the power
This line:

\$P_1 + P_2 + P_3 + \frac{v^2}{\infty} = 570\$

is not valid, because total power changes as there is less loads (resistors) in the circuit. 

You can see this by calculating the equivalent resistance, in this case was \$\frac{R}{4}\$ and now is \$\frac{R}{3}\$, so the resistance is higher and decreases the power.

Actually each resistor in parallel is responsible for a part of the total power, so, before, you had:

\$P_1 = P_2 = P_3 = P_4 = \frac{570}{4} = 142,2\$

So as long as voltage remains the same over the resistors, dissipated power will remain the same.

Now you have only three resistors, so:

\$P_{total} = P_1 + P_2 + P_3 = 3*142,2 = 427,5\$