In the proof of Lemma $3.36$ in Algebraic Geometry and Arithmetic Curves, it is stated that, if $B=\oplus_{d\ge0}B_d$ is a graded algebra over a ring $A,$ and if $I$ is an ideal of $B,$ then $$V(I)\cap\operatorname{Proj}(B)=V_+(I^h),$$ where $I^h=\oplus_{d\ge0}I\cap B_d$ is the homogenized ideal of $I,$ thus the topology on $\operatorname{Proj} B$ is induced from that on $\operatorname{Spec}(B).$ The containment of the left side in the right one being clear, I don't see why this is an equality.
I have no idea how can one be sure that a prime ideal in $\operatorname{Proj}B$ contains $I$ iff it contains $I^h.$
Any hint or reference is well welcomed.
Edit: I thought that this came from the equality $$\sqrt I=\sqrt{I^h},$$ which is false, thanks to a comment by @user121097.
I changed the question according to the quoted comment, sorry for this.
P.S.: The title does not match the question exactly. Apology again.
Edit: As pointed out in the comments below, I was not using the correct definition of $I^h$. I do recall working with this construction some time ago: it is actually the largest homogeneous ideal contained in $I$. An interesting fact that may be useful is that, if $P$ is prime in $B$, then $P^h$ is a homogeneous prime ideal. If a solution to the actual question comes to me, I will of course let you know.
Old, incorrect answer:
This is clear once you realize that $I^h$ is the smallest homogeneous ideal containing $I$. Thus if $P$ is a homogeneous ideal, then $I \subseteq P \iff I^h \subseteq P$ (the reverse implication following from $I \subseteq I^h$). In my opinion, the equality follows easily from this.
To explicitly show that $V_+(I^h) \subseteq V(I)\cap\operatorname{Proj}(B)$ as you want, suppose that $P \in \operatorname{Proj}(B)$ with $I^h \subseteq P$. Then $I \subseteq I^h \subseteq P$, so $P \in V(I) \cap \operatorname{Proj}(B)$.