Using the [Markov condition](https://en.wikipedia.org/wiki/Causal_Markov_condition), the conditional dependencies in your DAG are:
\begin{align}
F &\perp P \mid \emptyset \\
C &\perp P \mid F \\
L &\perp \{F,D\} \mid C \\
D &\perp \{F,P,L\} \mid C \\
P &\perp \{C,F,D\} \mid \emptyset
\end{align}
Given the joint probability distribution $p(C,F,P,D,L)$, suppose you compute
$$
p(F,P), \ p(F), \ p(P) \\
p(C,P\mid F), \ p(C\mid F), \ p(P \mid F) \\
p(L,F,D\mid C), \ p(L\mid C), \ p(F,D \mid C) \\
p(D,F,P,L\mid C), \ p(D\mid C), \ p(F,P,L \mid C) \\
p(P,C,F,D), \ p(P), \ p(C,F,D)
$$
for all values of $C,F,P,D,$ and $L$. If you find that
\begin{align}
p(F,P) &= p(F) \cdot p(P) \\
p(C,P\mid F) &= p(C\mid F) \cdot p(P \mid F) \\
p(L,F,D\mid C) &= p(L\mid C) \cdot p(F,D \mid C) \\
p(D,F,P,L\mid C) &= p(D\mid C) \cdot p(F,P,L \mid C) \\
p(P,C,F,D) &= p(P) \cdot p(C,F,D)
\end{align}
for all values of $C,F,P,D,$ and $L$, then the DAG is an I-map for the probability distribution $p(C,F,P,D,L)$. In other words, if you can show that **all** conditional independencies encoded by the DAG are encoded in $p(C,F,P,D,L)$, then the DAG is an I-map for $p(C,F,P,D,L)$.

Note, however, that there may be other conditional independencies encoded in $p(C,F,P,D,L)$ that are not represented in the DAG. If it so happens that **all** conditional independencies encoded in $p(C,F,P,D,L)$ are also represented in the DAG, then the DAG is a [perfect map](https://ermongroup.github.io/cs228-notes/representation/directed/) for $p(C,F,P,D,L)$.