Can you draw its graph ? -1

The Answer given here is wrong. n(A) = 0 Therefore, n ( P(A) ) = 2^0 = 1 And so, n ( P ( P(A) ) ) = 2^1 = 2

$x^2-4x+\lfloor{x}\rfloor+3=0$ $x^2-4x+x-\left\{ x \right\} +3=0$ $x^2-3x +3=\left\{ x \right\}$

Now drawing graphs for them

As here we can see graph of quadratic eq intersects with $\left\{ x \right\}$ at $x=1,2$ but at these x we get $y=1$ but $0 \le \left\{ x \right\}<1$ hence there is no solution to this equation.

$A \equiv \{\phi \}$

Hence $\boxed {n(P(P(A)))=4}$

For bigger detailed graph check this link

This can also be done algebraically as follows:

$x^2-3x +3=\left\{ x \right\}$ $0\le x^2-3x +3<1$ $x \in (1,2)$

We know $\left\{ x \right\}$ represents straight lines such that: $y=x+c$ for $0\le y <1$ and here $c \in \{-1,0,1,2.......\}$ .

Now the lines $y=x+c$ , for $c \in \{-1,0\}$ intersects the quadratic equation at $(1,1),(2,1)$ but $\left\{ x \right\} <1$ hence No solution .

Rest of the lines intersect at pts for which ordinate $>1$ hence no solution for them obviously

So $n(P(P(A)))$ =4

Note : $y=x-1$ is a tangent to quadratic curve and $y=x$ intersects at 2 pts but i have taken only 1 pt i.e $(1,1)$ Coz after that $y>1$ and $y=1$ also is not in range of fraction part of x.

A={2}; p(A)= $2^1$ p(p(A))= $2^2 = 4$

A={} so P(P(A)) is 2^2.