well this is a vert interesting question and let me help u out with it

 

well the 10^(-8) M solution of NaOH is a very dilute one and we need to take into account the OH^-  ions coming from water also into consideration ..

 

so total concentration of OH^-  ions is 10^(-8) + 10^(-7)

 

and thus the pOH is -log(10^(-8) + 10^(-7))  = 6.96

 

and thus pH is 14 - 6.96 = 7.04

 

cheers