Here is what I finished with, although the problem states that it is a tautology and not a contingency.
$For :(A \lor B) \land (A \rightarrow C) \lor (B \rightarrow D) \rightarrow (C \lor D)$
$W(A/true)$ $ = (True \lor B) \land (True \rightarrow C) \lor (B \rightarrow D) \rightarrow (C \lor D)$ --- First condition
$\equiv True \land (True \rightarrow C) \lor (B \rightarrow D) \rightarrow (C \lor D)$ ~~~~~~ $True \lor B \equiv True$
$\equiv True \land (C) \lor (B \rightarrow D) \rightarrow (C \lor D)$ ~~~~~~ $True \rightarrow C \equiv C$
$\equiv C \lor (B \rightarrow D) \rightarrow (C \lor D)$ ~~~~~~ $True \land C \equiv C$
$X1(D/True) $ $\equiv C \lor (B \rightarrow True) \rightarrow (C \lor True)$ ~~~~~~ First condition
$\equiv C \lor (B \rightarrow True) \rightarrow True$ ~~~~~~ $C \lor True \equiv True$
$\equiv C \lor True \rightarrow True$ ~~~~~~ $B \rightarrow True \equiv True$
$\equiv True \rightarrow True$ ~~~~~~ $C \lor True \equiv True$
$\equiv True$ ~~~~~~ $True \rightarrow True \equiv True$
$X1(D/False) $ $ \equiv C \lor (B \rightarrow False) \rightarrow (C \lor False)$ ~~~ ~~~ First condition
$\equiv C \lor B \rightarrow (C \lor False)$~~~~~~$ B \rightarrow False \equiv \neg B$
$\equiv C \lor B \rightarrow C$~~~~~~$ C \lor False \equiv C$
$\equiv C \rightarrow C$~~~~~~$ C \lor B \rightarrow C \equiv C \rightarrow C$
$\equiv True$~~~~~~$ C \rightarrow C \equiv True$
$W(A/False)$ $= (False \lor B) \land (False \rightarrow C) \lor (B \rightarrow D) \rightarrow (C \lor D)$ ~~~ First condition
$\equiv B \land (False \rightarrow C) \lor (B \rightarrow D) \rightarrow (C \lor D)$~~~~~~$ False \lor B \equiv B$
$\equiv B \land True \lor (B \rightarrow D) \rightarrow (C \lor D)$~~~~~~$ False \rightarrow C \equiv True$
$\equiv B \lor (B \rightarrow D) \rightarrow (C \lor D)$~~~~~~$ B \land True \equiv B$
$X2(D/True) $ $ = B \lor (B \rightarrow True) \rightarrow (C \lor True)$ ~~~ ~~~ First Condition
$\equiv B \lor True \rightarrow (C \lor True)$~~~~~~$ B \rightarrow True \equiv True$
$\equiv (B \lor True) \rightarrow True$~~~~~~$ C \lor True \equiv True$
$\equiv True \rightarrow True$~~~~~~$ B \lor True \equiv True$
$\equiv True$ Because $True \rightarrow True \equiv True$
$X2(D/False) $ $= B \lor (B \rightarrow False) \rightarrow (C \lor False)$ ~~~ ~~~ First Condition
$\equiv B \lor (\neg B) \rightarrow (C \lor False)$~~~~~~$ B \rightarrow False \equiv \neg B$
$\equiv True \rightarrow (C \lor False)$~~~~~~$ B \lor (\neg B) \equiv True$
$\equiv True \rightarrow C$~~~~~~$ C \lor False \equiv C$
$\equiv C$~~~~~~$ True \rightarrow C \equiv C$
Contingency!?
It is supposed to come out as a tautology. I'm using Quine's method of substitution.
Edit: cleaned up