Let $h:Y \to B$ be a surjective Serre fibration and let the following be a pullback diagram. $$\require{AMScd} \begin{CD} X @>>> E @. \\ @VfVV @VgVV \\ Y @>>h> B @. \end{CD}$$ Then $g$ is a Serre fibration iff $f$ is one.
Showing that $g$ Serre fibration implies $f$ Serre fibration is just a simple application of the pullback property and doesn't use any property of $h$.
I'm struggling with proofing the other direction. I tried the following:
In order to show that $p:E\to B$ is a Serre fibration, consider the commutative diagram $$\require{AMScd} \begin{CD} D^n @>>> E @. \\ @VVV @VgVV \\ D^n \times I @>k>> B. @. \end{CD}$$We now need to show that there exists a lift $D^n\times I \to E$.
If I could show that there exists a continuous map $D^n \to Y$, such that the diagram
$$\require{AMScd} \begin{CD} D^n @>>> Y @. \\ @VVV @VhVV \\ D^n\times I @>k>> B. @. \end{CD}$$
commutes, we could use the lifitng prop of the Serre fibration $h$ and the pullback property to induce a comm. diagram
$$\require{AMScd} \begin{CD} D^n @>>>X @>>> E @. \\ @VVV @VfVV @VgVV \\ D^n\times I @>>>Y @>>h> B @. \end{CD}$$ and use the lifting prop. of the Serre fibration $f$ to get a lift $D^n\times I \to E$. Without such a map
Can I assure that there is such a continuous map $D^n \to Y$?
Using surjectivity I get such a map in set, but this isn't necessary continuous. Without such a map, I don't see how I could proof this statement.