By recursively applying the recurrence relation, we got
T(n) = $2^k . T(n^{\frac{1}{2^k}}) + k . lg n$
By taking Base condition, T(2) = 1 ( i assumed it, in the question they didn't specify it )
$n^{\frac{1}{2^k}} = 2 ===> 2^k = log_2n \;\;and\;\; k = log_2(log_2n)$
∴ T(n) = $log_2n . T(2) + log_2(log_2n) . log_2n$
∴ T(n) = $log_2n + log_2(log_2n) . log_2n = lg \;n + lg\;n . (lg\;lg\;n) = O(lg\;n . (lg\;lg\;n) ) $