How to determine potential difference of a point on a circuit relative to another point?
Assume wires have zero resistance, and take the negative battery terminal to be the 0v reference:

You therefore know that point "A" must be at +6v.

Assume that the diode will clamp the voltage across it at 0.7v when forward biased, and will have infinite resistance when negative biased.

Therefore, point "B" cannot be at a higher voltage than A + 0.7, but can be any amount lower. The point at which the circuit is in equilibrium can then be determined by calculating the voltage of the potential divider formed by the 200 ohm and 1000 ohm resistors, and (in a sense) seeing how the diode would react. (The 1200 ohm resistor has no bearing on the calculations: since its terminals are fixed at +6v and 0v and cannot affect anything else in the circuit).

So, without the diode, the current through the 1000 and 200 can be used to calculate the voltage at B. The current through one is equal to the current through the other. Since resistors in series are equivalent to the resistance of a resistor with the value of their sum, the current is equal to:

I = V / R = 6 / (200 + 1000) = 5mA

Now you can calculate the voltage across each individually:

V = I R

Therefore, the voltage across the 200 ohm resistor is:

V = 200 x 0.005 = 1v

And across the 1000 ohm resistor:

V = 1000 * 0.005 = 5v.

So ignoring the diode, the voltage at B would be 1v.

Now, with the diode, we said it is effectively open circuit if B is below A. Which we have now established it is. Therefore, the diode in this circuit is not doing anything, and is equivalent to an open circuit. Therefore, B will be at 1v.