Map $p:E\to B$ is a fibration iff map from a pullback has a section Given $p:E\to B$ I have space $P = \{(e, \gamma)\in E \times B^I | p(e) = \gamma (0)\}$. I'm trying to show that the map $p$ is a fibration iff $q:E^I\to P, \omega \mapsto (\omega (0),p\circ \omega$) has a section $s:P\to E^I$.
I think that P is a pullback of $E\to B \leftarrow B^I$ which gives us a unique map to $P$ but I don't see that this helps us to give a section, and we don't have a homotopy to lift. Maybe we try to lift the path from a singleton?

You have made the key observation that $P$ is the pullback of $E\to B\leftarrow B^I$. Next note that a homotopy $f: X\times I\to B$ corresponds to a map $X\to B^I$, which I'll abuse the notation and also call $f$. There is also the lift of $f|_{X\times\{0\}}$, which I will call $\tilde{h}_0$. Thus you can form a pullback cone from $X$ to $E\to B\leftarrow B^I$. (See the diagram below.)

The fact that this cone out of $X$ commutes depends on the definition of the map $B^I\to B$, which evaluates every path at $0$, so the composition $X\to B^I\to B$ is precisely $f|_{X\times\{0\}}$, i.e. the "starting map" of the homotopy $f$.
Now you have a factoring through the pullback $P$. Composing the factoring map $h: X\to P$ with the section $s$ gives you a map $X\to E^I$, which corresponds the desired lifting.

The converse follows by applying the homotopy lifting property to $P\to B^I$ and $P\to E$. Essentially, we put $P$ in the place of $X$ in the above diagram.

The left map $h: P\to B^I$ defines a homotopy out of $P$. The top map $\tilde{h}_0$ is a lifting of the "starting map". The homotopy lifting property then gives us a lifting $s: P\to E^I$. This is our section.
Finally you can check that the fact that $s$ is a lifting of $h$ that extends $\tilde{h}_0$ guarantees that $q\circ s$ factors the outer pullback cone, but since the identity map on $P$ also satisfies this property, we must have $q\circ s = \operatorname{id}_P$.
