I have modeled a rigid (Dia = Branch Dia) from the centre of the Header to Surface and I started my Branch from the end of rigid element as shown and given user defined SIF in nodes 1600 and 1595. Does CAESAR II apply Class 1 branch flexibility here, Since my start of the branch is a rigid element?
Please refer the below image for further details.

Try running it with and without the rigid element and compare the results.

Edit to add:

I didn't make out the light gray lines on the yellow background the first time I looked at the image.

For laterals, it is recommended to model it as you have shown, except with CNODEs between the rigid element and the branch, and to apply flexibilities as calculated through FEA software, such as Nozzlepro.

Thanks for your reply Sir..
For a rigid element, the stiffness properties are computed using 10times of the entered thickness.

While calculating Kx and Kz using Class 1 branch flexibility, which thickness will be used for the above case.
Entered thickness
OR
10times the entered thickness.
An extract from User's guide is shown below..
Thanks in advance

To consider class one branch flexibilities, you need model branch connection as per Normal practice. No need to model rigid element. CAESAR II will consider flexibilities if applicable as per criteria. You just need to activate class one flexibilities in configuration set up.