Let $k = \mathbb{R}$ and let $P$ be the prime ideal of $\mathbb{R}[x,y]$ generated by $h := y^4 + y^2 + (x^2-1)^2$. According to Sage¹ (though I'm sure this is easy to check by other means), $h$ is irreducible in $\mathbb{C}[x,y]$, so certainly in $\mathbb{R}[x,y]$, so $P$ is prime. But $h(x,y)=0$ over $\mathbb{R}$ can only occur for $(x,y) = (\pm 1,0)$ for obvious reasons of positivity, and indeed does happen for those two points. And $V = \{(1,0),(-1,0)\}$ certainly isn't irreducible, nor is the ideal $I(V)$ prime.
Geometrically, $\{h=0\}$ is an irreducible curve with exactly two singular points (ordinary double points), which are its only two real points.
- Actually checking over the algebraic closure of $\mathbb{Q}$:
sage: R.<x,y> = PolynomialRing(QQbar,2)sage: factor(y^4 + y^2 + (x^2-1)^2)x^4 + y^4 + (-2)*x^2 + y^2 + 1