1. Simplify the expression inside the limit

We are given

$$p=\lim_{x\to 0^+}\left(1+\tan^2\sqrt{x}\right)^{\frac{1}{2x}}.$$

Using the identity $1+\tan^2\theta=\sec^2\theta,$ we get

$$p=\lim_{x\to 0^+}\left(\sec^2\sqrt{x}\right)^{\frac{1}{2x}}.$$

So,

$$p=\lim_{x\to 0^+}\left(\sec \sqrt{x}\right)^{\frac{1}{x}}.$$

2. Take logarithm

Let $\log p = \lim_{x\to 0^+} \frac{1}{2x}\log\left(1+\tan^2\sqrt{x}\right).$

Since $1+\tan^2\sqrt{x}=\sec^2\sqrt{x}$,

$$\log p=\lim_{x\to 0^+}\frac{1}{2x}\log(\sec^2\sqrt{x}) =\lim_{x\to 0^+}\frac{1}{x}\log(\sec\sqrt{x}).$$

3. Use standard expansion near $0$

For small $t$,

$$\cos t = 1-\frac{t^2}{2}+o(t^2).$$

Hence,

$$\sec t = \frac{1}{\cos t}=1+\frac{t^2}{2}+o(t^2).$$

Therefore,

$$\log(\sec t) \sim \frac{t^2}{2} \quad (t\to 0).$$

Now put $t=\sqrt{x}$. Then $t^2=x$, so

$$\log(\sec \sqrt{x}) \sim \frac{x}{2}.$$

Thus,

$$\log p = \lim_{x\to 0^+}\frac{1}{x}\cdot \frac{x}{2}=\frac{1}{2}.$$

4. Find the correct option

Hence,

$$\boxed{\log p=\frac{1}{2}}.$$

So the correct option is $\boxed{\text{A}}.$

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.