Ideal inductor in series with a resistor:
But, how can that happen it would leave no voltage for the
  resistance??


Andy's given the correct answer to this question in the comments.  This is the complete analysis.

Call the left winding primary and the right winding secondary.  See that the secondary winding is open and thus there is zero secondary current.

Since this is an ideal transformer, the primary current must also be zero and thus, by Ohm's Law, the voltage across the \$2 \Omega\$ resistor, in series with the primary, is zero.

By KVL, the voltage across the primary is \$-1V\$ so the voltage across the secondary is \$-4V\$.

Finally, the open circuit voltage is \$V_{ab} =V_{th} = 1 - 4 = -3V = 3\angle 180 V\$

To find the Thevenin impedance directly, zero the independent voltage source and see that the \$2 \Omega\$ resistor is across the primary which, reflected to the secondary, appears as \$2 \cdot 4^2 = 32 \Omega\$.  Thus, the Thevenin Impedance is \$Z_{th} = 32 \Omega\$.