How to distribute $(\neg  \lor ) \land [((\neg  \lor ) \land ( \lor \neg ))∨((\neg  \lor ) \land (\lor \neg ))]?$ I see two possible ways to distribute this. 
I believe it is case 1.
For brevity,
Let 
\begin{align}A &= (\neg P \lor Q) &
 B &= (P \lor \neg Q) & C &=(\neg R \lor Q) & D &=(R \lor \neg Q).
\end{align}
Case 1:
$$(\neg P \lor R) \land [(A\land B)\lor (C\land D)]=$$
$$[(\neg P\land A \land B)\lor(\neg P \land C \land D)]\lor [(R \land A \land B)\lor (R \land C \land D)]$$
Case 2:
$$(\neg P \lor R)\land [(A \land B)\lor (C\land D)]=$$
$$[(\neg P \lor A) \land (\neg P \land B)\lor(\neg  \land C)∧(\neg  \land D]\lor [(R \land A)\land (R \land B)) \lor (R \land C) \land (R \land D)]$$
Question: Which case is correct (if either)? Why is this case correct/other wrong?
 A: Case 1 is correct.
What you end up with in case 2 is ungrammatical ... you'll need to add some parentheses.  I assume you meant:
$$[\color{red}((\neg P \color{red}\land A) \land (\neg P \land B)\color{red})\lor\color{red}((\neg  \land C)∧(\neg  \land D\color{red})\color{red})]\lor [\color{red}((R \land A)\land (R \land B)) \lor \color{red}((R \land C) \land (R \land D)\color{red})]$$
But note that that is equivalent to:
$$[((\neg P \land A) \land (\neg P \land B))\lor((\neg  \land C)\land(\neg  \land D))]\lor [((R \land A)\land (R \land B)) \lor ((R \land C) \land (R \land D))] \overset{Association \ (drop \ parentheses \ in \ generalized \ conjunctions)}\Leftrightarrow$$
$$[(\neg P \land A \land \neg P \land B)\lor(\neg  \land C\land \neg  \land D)]\lor [(R \land A\land R \land B) \lor (R \land C \land R \land D)] \overset{Commutation}\Leftrightarrow$$
$$[(\neg P \land \neg P \land A \land B)\lor(\neg  \land \neg  \land C \land D)]\lor [(R \land R \land A \land B) \lor (R \land R \land C \land D)] \overset{Idempotence}\Leftrightarrow$$
$$[(\neg P \land A \land B)\lor(\neg  \land C \land D)]\lor [(R \land A \land B) \lor (R \land C \land D)]$$
... which is exactly what you got from case 1
In other words, had you used proper parentheses, both results would have been correct!
That said, I have a feeling you did the following to get to case 2:
$$(\neg P \lor R) \land [(A\land B)\lor (C\land D)]=$$
$$[(\neg P \land ((A\land B)\lor (C\land D))] \lor [(R \land ((A\land B)\lor (C\land D))]=$$
$$[(\neg P \land (A\land B)) \lor (\neg P \land (C\land D))] \lor [(R \land (A\land B))\lor (R \land (C\land D))]=$$
$$[(\neg P \land A) \land (\neg P \land B)) \lor (\neg P \land C) \land (\neg P \land D))] \lor [(R \land A) \land (R \land B))\lor (R \land C) \land (R \land D))]$$
That is, you acted as if that last step was a Distribution ... but it isn't! ... because you end up distributing a $\land$ over a $\land$.  Distribution is $\land$ over $\lor$ or vice versa.
So, you should have just stopped 1 line earlier, and thus have ended up with 
$$[(\neg P \land (A\land B)) \lor (\neg P \land (C\land D))] \lor [(R \land (A\land B))\lor (R \land (C\land D))]$$
but since $\land$ is Associative you can drop some parantheses and thus get:
$$[(\neg P \land A \land B)\lor(\neg  \land C \land D)]\lor [(R \land A \land B) \lor (R \land C \land D)]$$
and since $\lor$ is Associative you can drop those square brackets as well:
$$(\neg P \land A \land B)\lor(\neg  \land C \land D)\lor (R \land A \land B) \lor (R \land C \land D)$$
A: Case 1 is right. Case 2 is not even a formula since some parentheses are missing: in $(\lnot P \lor A) \land (\lnot P \land B) \lor (\lnot P \lor C) \land (\lnot P \lor D)$ it is not clear what is the principal connective.
Let us see why case 1 is correct. Essentially, you use repeatedly the following distributive law (together with associativity and commutativity):
\begin{align}
 A \land (B \lor C) &\equiv (A \land B) \lor (A \land C) 
\end{align}
By applying step-by-step this equivalence rule, you get:
\begin{align}
  &(\neg P \lor R) \color{red}{\land} \big((A\land B) \color{red}{\lor} (C\land D)\big) \\
\text{(distributivity) } \ \equiv \ & ((\neg P \lor R) \land (A \land B)) \lor ((\neg P \lor R) \land (C \land D)) \\
\text{(commutativity) } \ \equiv \ & ((A \land B) \color{red}{\land} (\neg P \color{red}{\lor} R)) \lor ((C \land D) \color{red}{\land} (\neg P \color{red}{\lor} R)) \\
\text{(distributivity) } \ \equiv \ & \big(((A \land B) \land \neg P) \lor ((A \land B) \land R)\big) \lor \big(((C \land D) \land \lnot P) \lor ((C \land D) \land R)\big) \\
\text{(associativity) } \ \equiv \ & \big((A \land B \land \neg P) \lor (A \land B \land R)\big) \lor \big((C \land D \land \lnot P) \lor (C \land D \land R)\big)
\end{align}
