or we can simply see like this,
$(log log n)! $ we can write lik this =$(log log n)^{log log n}$ ,suppose $(log log n)$=$y$
a.$y^y$=taking log $\rightarrow $ y log y
b.$y^{log \ n}$=taking log $\rightarrow $$ logn \ y$
c.$y^{log \ y}$=taking log $\rightarrow $ log y.y
d.$log n ^{ y}$=taking log $\rightarrow $ y.log log n
e.$2^\sqrt{y}$=taking log $\rightarrow $$\sqrt{y}$.log 2
when put value of y then we can see only option (b) is greater than all others because it has log n . others does not have simple log n