For a prime number $p$, there exists $n\in\mathbb{Z}_+$, $\sqrt{p+n}+\sqrt{n}$ is an integer, then
$\text{(A)}$ there is no such $p$
$\text{(B)}$ there in only one such $p$
$\text{(C)}$ there is more than one such $p$, but finitely many
$\text{(D)}$ there are infinitely many such $p$
1. We start by analyzing the given condition: $\sqrt{p+n} + \sqrt{n}$ is an integer. Let this integer be denoted by $k$. Therefore, we can write:
   \[
   \sqrt{p+n} + \sqrt{n} = k
   \]
2. Rearrange the equation to isolate $\sqrt{p+n}$:
   \[
   \sqrt{p+n} = k - \sqrt{n}
   \]
3. Square both sides to eliminate the square roots:
   \[
   p + n = (k - \sqrt{n})^2
   \]
4. Expand the right-hand side:
   \[
   p + n = k^2 - 2k\sqrt{n} + n
   \]
5. Subtract $n$ from both sides:
   \[
   p = k^2 - 2k\sqrt{n}
   \]
6. For $p$ to be an integer, the term $2k\sqrt{n}$ must also be an integer. This implies that $\sqrt{n}$ must be an integer. Let $\sqrt{n} = m$ where $m$ is an integer. Therefore, $n = m^2$.
7. Substitute $n = m^2$ back into the equation:
   \[
   p = k^2 - 2km
   \]
8. Since $p$ is a prime number, we need to find conditions under which $k^2 - 2km$ is a prime number. Rewrite the equation:
   \[
   p = (k - m)^2 - m^2
   \]
9. Let $b = k - m$. Then:
   \[
   p = b^2
   \]
10. Since $p$ is a prime number, $b^2$ must be a prime number. However, the only prime number that is a perfect square is $2$, which is not possible since $2$ is not a perfect square. Therefore, $b$ must be an integer such that $b^2$ is a prime number, which is impossible.

11. Therefore, there are no such prime numbers $p$ that satisfy the given condition.

Conclusion:
\[
\boxed{\text{(A)}}
\]