Using the 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 for $p(C,F,P,D,L)$.

Using the Markov condition, the conditional independencies 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 for $p(C,F,P,D,L)$.

Using the 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 the values $$ p(F,P), \ p(F), \ p(P) \\ p(C,P\mid F), \ p(C\mid F), \ p(P \mid F) \\ p(L,F,D\mid C), \ p(L\mid C), \ p(F,D \mid C) \\ p(D,F,P,L\mid C), \ p(D\mid C), \ p(F,P,L \mid C) \\ p(P,C,F,D), \ p(P), \ p(C,F,D) \\ $$ 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 for $p(C,F,P,D,L)$.

Using the 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 for $p(C,F,P,D,L)$.