This is a $1^\infty$ indeterminate form. Take logarithm:

$\ln p = \lim_{x\to0^+}\dfrac{\ln(1+\tan^2\sqrt{x})}{2x}$

Let $t=\sqrt{x}$, so $x=t^2$, $x\to0^+$ means $t\to0^+$:

$= \lim_{t\to0^+}\dfrac{\ln(1+\tan^2 t)}{2t^2} = \lim_{t\to0^+}\dfrac{\tan^2 t}{2t^2} = \dfrac{1}{2}$

(using $\ln(1+u)\approx u$ for small $u$ and $\lim_{t\to0}\frac{\tan t}{t}=1$)