Let $(E,\mathcal E,\mu)$ be a finite measure space, $p\ge1$ and $\varphi\in L^p(\mu)'$. Note that $$\nu(B):=\varphi(1_B)\;\;\;\text{for }B\in\mathcal E$$ is a well-defined real-valued measure on $(E,\mathcal E)$ with $\nu\ll\mu$. Hence, $$\nu=f\mu$$ for some $f\in L^1(\mu)$ by the Radon-Nikodým theorem.
Does it somehow follow from $\varphi\in L^p(\mu)'$ that $f\in L^p(\mu)$?
It's standard that $L^{p}(\mu)'$ is isomorphic to $L^{q}(\mu)$, where $q \geq 1$ is the dual exponent satisfying $q^{-1} + p^{-1} = 1$. Specifically, the map $\Phi : L^{q}(\mu) \to L^{p}(\mu)'$ given by \begin{equation*} [\Phi(f)](g) = \int_{E} g(x) f(x) \, \mu(dx) \end{equation*} is linear, continuous, and bijective. This is proved in textbooks on real analysis, often under the name "Riesz Representation Theorem."
Hence, in your case, we can find $f \in L^{q}(\mu)$ so that $\varphi = \Phi(f)$. Notice $f$ need not be in $L^{p}(\mu)$ (if $p > 2$).