Applying De Morgan's Law, we get:
\[
eg (p \vee q) = eg p \wedge eg q
\]
Therefore:
\[
(eg p \wedge eg q) \vee (eg p \wedge q)
\]
Using the distributive law:
\[
eg p \wedge (eg q \vee q)
\]
Since \( eg q \vee q = 1 \) (a tautology):
\[
eg p \wedge 1 = eg p
\]
The Boolean expression simplifies to:
\[
eg p
\]