Let $x$ : An object
$h(x)$: $x$ is a house and
$b(x)$: $x$ is a brick
Houses are not bricks $\equiv$ For all $x$ , if $x$ is a house then it is not a brick $\equiv \forall x( h(x) \rightarrow \sim b(x)) \equiv \forall x (\sim h(x)\ \vee \sim b(x))$
This above proposition is considered false so will negate it.
$\sim( \forall x (\sim h(x)\ \vee \sim b(x))) \equiv \exists x ( h(x)\ \wedge b(x)) $
- All houses are bricks $\equiv$ For all $x$ , if $x$ is a house then it is also a brick $\equiv \forall x ( h(x) \rightarrow b(x)) \equiv \forall x (\sim h(x)\ \vee b(x))$
- No house is a brick $\equiv$ There does not exist an $x$ such that $x$ is a house and it is a brick $\equiv \sim \exists x ( h(x) \wedge b(x)) \equiv \forall x (\sim h(x)\ \vee \sim b(x) )$
- Some houses are bricks$\equiv$ There exists some $x$ such that $x$ is a house and it is a brick $\equiv \exists x(h(x) \wedge b(x))$
- Some houses are not bricks$\equiv$ There exists some $x$ such that $x$ is a house and it is not a brick $\equiv \exists x(h(x) \wedge \sim b(x))$
$\therefore$ Option $D.$ is the correct answer.