Given:

Top gate (NOR):

$X=\overline{A+B}$

Bottom gate (NAND):

$Z=\overline{AB}$

Final output:

$Y=X\cdot Z$

Substituting the values of $X$ and $Z$:

$Y=\overline{A+B}\cdot \overline{AB}$

Using De Morgan's Theorem:

$\overline{A+B}=\overline{A}\overline{B}$

$\overline{AB}=\overline{A}+\overline{B}$

Therefore,

$Y=\overline{A}\overline{B}\left(\overline{A}+\overline{B}\right)$

Applying the absorption law:

$Y=\overline{A}\overline{B}$

Hence,

$\boxed{Y=\overline{A}\overline{B}}$

or equivalently,

$\boxed{Y=\overline{A+B}}$ $\to$ NOR Gate