As your testing H0:p=0.79, then you have a two-tailed test.
The p-value at two-tailed test is given by:
Then, you need to check into the standard normal cumulative table to find at which -z you do have a probability of 0.395. Thus:
As you can see in the picture at z=-0.26 you have a P(Z<=-z)=0.3974
And at z=-0.27 you have a P(Z<=-z)=0.3936.
The average of these values is:
Then, you have a probability of 0.395 at a z of:
Then -z=-0.27, thus z=0.27
