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You can verify this equivalence elementarily (without the language of fibrations etc.):

Assume first that $C$ is the category with only one morphism (i.e. the terminal category), so that a presheaf on it is just a set. Then the statement is as follows: If $P$ is a set, then a set $F$ together with a map $F \to P$ is the same as to give a family of sets indexed by $P$. But this is obvious, right? Given $F \to P$, we may look at its fibers $F_s$, where $s$ runs through the elements $s \in P$. Since $F$ is the disjoint union of the $F_s$, it is also clear how to write down the inverse.

For general $C$ it works in the same way, $C$ is just a sort of parametrization.

If $F \in \widehat{C} / P$, i.e. $F$ is a presheaf on $C$ together with a morphism $F \to P$, then define a presheaf $G$ on $\int_C P$ as follows: If $(X,s)$ is an element of $P$, i.e $s \in P(X)$, then let $G(X,s)$ be the fiber of $s$ with respect to $F(X) \to P(X)$.

Conversely, if $G$ is a presheaf on $\int_C P$, then define a presheaf $F$ on $C$ as follows: For $X \in C$ let $F(X) = \coprod_{s \in P(X)} G(X,s)$. We have a natural projection $F(X) \to P(X)$, which gives rise to a morphism $F \to P$.

Now it is straight forward to check that these assignments actually define functors which are pseudo-inverse to each other.

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You can verify this equivalence elementarily (without the language of fibrations etc.):

If $F \in \widehat{C} / P$, i.e. $F$ is a presheaf on $C$ together with a morphism $F \to P$, then define a presheaf $G$ on $\int_C P$ as follows: If $(X,s)$ is an element of $P$, i.e $s \in P(X)$, then let $G(X,s)$ be the fiber of $s$ with respect to $F(X) \to P(X)$.

Conversely, if $G$ is a presheaf on $\int_C P$, then define a presheaf $F$ on $C$ as follows: For $X \in C$ let $F(X) = \coprod_{s \in P(X)} G(X,s)$. We have a natural projection $F(X) \to P(X)$, which gives rise to a morphism $F \to P$.