Let $x = 2^k$

$T(x) = 3T\left(\frac{x}{2}\right) + 1$

We can apply Master's theorem case 1 with $a = 3$ and $b = 2$ as $f(x) = 1 = O\left(x^{\log_2 3 - \epsilon} \right), \epsilon > 0 $

So, $T(x) = \Theta \left(x^{\log_2 3}\right) \\= \Theta \left( {2^k}^{\log_2 3} \right)\\= \Theta \left({2^{\log_2 3}}^k \right)\\ = \Theta \left(3^k \right)$

So, only option possible is **B**.

We can also directly solve as follows:

$T(x) = 3T\left(\frac{x}{2}\right) + 1$

$\\\quad= 9T \left (\frac{x}{4}\right) + 1 + 3$

$ \\\quad \vdots$

$ \\\quad= 3^{\log_2 2^k} + \left( 1 + 3 + 9 + \dots + 3^{\log_2 {2^k-1}}\right)$

$\\\quad \left(\text{recursion depth is }\log_2 x \text{ and } x = 2^k\right)$

$ \\\quad= 3^{k} + \frac{3^{\log_2 {2^k}} - 1}{3-1}$

$ \\\quad \left(\text{Sum to n terms of GP with } a = 1 \text{ and } r = 3 \right)$

$\\\quad =3^k + \frac{3^k - 1}{2}$

$\\\quad=\frac{3. 3^k - 1}{2}$

$\\\quad=\frac{3^{k+1} -1}{2} $

OR

$T\left(2^k\right) = 3T\left(2^{k-1}\right) + 1$

$ \\\quad= 3^2T\left(2^{k-2}\right) + 1 +3$

$ \quad\vdots$

$ \\\quad= 3^k T\left(2^{k-k}\right) + \left( 1 + 3 + 9 + \dots + 3^{k-1}\right)$

$ \\\quad \left(\text{recursion depth is }k\right)$

$\\\quad= 3^k + \frac{3^{k -1}} {3-1}$

$\\\quad\left(\text{Sum to n terms of GP with } a = 1 \text{ and } r = 3 \right)$

$\\\quad=3^k + \frac{3^k -1}{2}$

$\\\quad=\frac{3. 3^k - 1}{2}$

$\\=\frac{3^{k+1} -1}{2} $