Solution:

The answer is 25. Let us consider the rectangle that forms the floor plan of the living room, and orient it so that the outlet $P$ lies on the horizontal side $AB$, dividing it into two segments of length $a$ and $b$. Let us also call $c$ the length of the two vertical sides $BC$ and $DA$, as in the figure. In this way, the area of the living room is $ac + bc$.

In order for the vertices $C$ and $D$ to be at distance less than or equal to $5$ from $P$, we must have $\sqrt{b^{2}+c^{2}} \leq 5$ and $\sqrt{a^{2}+c^{2}} \leq 5$, respectively. By a well-known inequality we have $ac \leq \frac{a^{2}+c^{2}}{2}$, with equality when $a=c$ is chosen, and similarly $bc \leq \frac{b^{2}+c^{2}}{2}$. Combining these inequalities we obtain that the area of $ABCD$ is
$$ac + bc \leq \frac{a^{2}+c^{2}}{2} + \frac{b^{2}+c^{2}}{2} \leq \frac{5^{2}}{2} + \frac{5^{2}}{2} = 25.$$
To verify that this is indeed the maximum area, we want to construct a configuration in which all these inequalities become equalities. We thus want in particular to have $a = c$ and $25 = a^{2} + c^{2} = 2a^{2}$, and solving we get $a = c = \frac{5}{\sqrt{2}}$. In the same way we obtain $b = c = \frac{5}{\sqrt{2}}$. Indeed, choosing $a = b = c = \frac{5}{\sqrt{2}}$, all points inside the rectangle are at distance from $P$ at most $\frac{5}{\sqrt{2}}$ both along the horizontal axis and along the vertical axis, and therefore have distance from it at most
$$\sqrt{\left(\frac{5}{\sqrt{2}}\right)^{2} + \left(\frac{5}{\sqrt{2}}\right)^{2}} = \sqrt{25} = 5$$

Second solution. Let us consider the points $C'$ and $D'$ obtained by reflecting $C$ and $D$ with respect to the side $AB$.

Then the rectangle $CD D' C'$ (which has double the area of $ABCD$) lies inside the circle with center $P$ and radius $5$. Therefore finding the maximum of the area of $ABCD$ is equivalent to constructing the rectangle of maximum area that lies inside a circle, and it is well known that this is the square. We then have $\overline{CD} = \overline{DD'} = \overline{D'C'} = \overline{C'C}$, and we conclude that $AD, BC$ must be half as long as $CD$.