An I-map is said to be perfect if $I(G)=I(p)$. Given a distribution $p$, it is not always possible to find a DAG $G$ such that $I(G)=I(p)$. Consider a joint distribution over four random variables such that $X_A \perp X_C|X_B,X_D$ and $X_B \perp X_D|X_A,X_C$ are the only conditional independence relations. One can verify that there is no four vertex DAG that implies only these conditional independence assumptions.
I am also aware that not even Markov networks are perfect maps for all probability distributions and so can't represent all conditional independencies. Why is this? I thought the whole point of probabilistic graphical models was to represent a joint distribution as efficiently as possible? If that is not possible, then how this problem overcome?