$T(n) = 2T\left(n^{\frac{1}{2}}\right)+1$
$ = 2\left(2T\left(n^{\frac{1}{2^2}}\right) + 1\right) + 1 $
$ = 4\times T\left(n^{\frac{1}{2^2}}\right) + 5 $
$ = 8\times T\left(n^{\frac{1}{2^3}}\right) + 13 \\ \cdots$
$=2^{(\lg \lg n)} + 2 \times \lg \lg n + 1\text{ (Proved below)} $
$= \Theta(\lg n)$
$n^{\frac{1}{2^k}} = 2 \\ \text{(Putting 2 so that we can take log.}\\\text{One more step of recurrence can't change the complexity.)} \\\implies \frac{1}{{2^k}} \lg n = 1 \text{(Taking log both sides)}\\\implies \lg n = 2^k \\\implies k = \lg \lg n$
So, answer is B, $T(n) = \Theta(\log n)$