Simplify $$$\overline{\left(\overline{A} + B\right) \cdot \left(\overline{B} + C\right)}$$$

Apply de Morgan's theorem $$$\overline{x \cdot y} = \overline{x} + \overline{y}$$$ with $$$x = \overline{A} + B$$$ and $$$y = \overline{B} + C$$$:

$${\color{red}\left(\overline{\left(\overline{A} + B\right) \cdot \left(\overline{B} + C\right)}\right)} = {\color{red}\left(\overline{\overline{A} + B} + \overline{\overline{B} + C}\right)}$$

Apply de Morgan's theorem $$$\overline{x + y} = \overline{x} \cdot \overline{y}$$$ with $$$x = \overline{A}$$$ and $$$y = B$$$:

$${\color{red}\left(\overline{\overline{A} + B}\right)} + \overline{\overline{B} + C} = {\color{red}\left(\overline{\overline{A}} \cdot \overline{B}\right)} + \overline{\overline{B} + C}$$

Apply the double negation (involution) law $$$\overline{\overline{x}} = x$$$ with $$$x = A$$$:

$$\left({\color{red}\left(\overline{\overline{A}}\right)} \cdot \overline{B}\right) + \overline{\overline{B} + C} = \left({\color{red}\left(A\right)} \cdot \overline{B}\right) + \overline{\overline{B} + C}$$

Apply de Morgan's theorem $$$\overline{x + y} = \overline{x} \cdot \overline{y}$$$ with $$$x = \overline{B}$$$ and $$$y = C$$$:

$$\left(A \cdot \overline{B}\right) + {\color{red}\left(\overline{\overline{B} + C}\right)} = \left(A \cdot \overline{B}\right) + {\color{red}\left(\overline{\overline{B}} \cdot \overline{C}\right)}$$

Apply the double negation (involution) law $$$\overline{\overline{x}} = x$$$ with $$$x = B$$$:

$$\left(A \cdot \overline{B}\right) + \left({\color{red}\left(\overline{\overline{B}}\right)} \cdot \overline{C}\right) = \left(A \cdot \overline{B}\right) + \left({\color{red}\left(B\right)} \cdot \overline{C}\right)$$