Push pull converter magnetizing current during both switch off state?
In a push pull converter during turn on a switch magnetizing current
  and load reflected current build up.


Correct


  So magnetizing inductance store energy.


Correct


  During both switch off state secondary coil does not carry current.


Incorrect. The residual energy in the primary is in the form of magnetism and that can equally be used in the secondary as it would be in a flyback converter so, one of the secondary half windings will be forward biased and push that energy to the load: -



Either (a) current will flow down the red path or (b) current will flow down the blue path and the flux will become reset.


  But when both switch off waveform shows primary current is zero. So my
  question is where does the charged magnetizing current discharges?


I think the above explains that.



Additional information

I've added a simulation to show what I mean: -



The circuit is this: -



And this substantiates what I was saying; D1 and D2 do not conduct the secondary flyback current at the same time. If M1 was pulsed on and off then D1 will conduct the flyback current to the load. If M2 is pulsed on and off then D2 conducts the flyback current to the load. And, during that flyback period (just after the pulse ends) the primary voltage doesn't fall to zero volts because it cannot - it stays at a voltage that is determined by the rate of change of the relevant diode (D1 or D2) current.

This is for a push pull converter with a coupling factor of 0.95 and 1:1:1:1 inductors.