Do only discs, sectors of discs, annuli, sectors of annuli and parallelograms have this property? The property I'm talking about is:

There is some partition of the plane figure P into $n$ congruent figures for any $n$.

Is it true that only discs, sectors of discs, annuli, sectors of annuli and parallelograms have this property?
 A: To expand on Hagen von Eitzen's remarks: the answer to your question is "no".
Let $a, b$ be positive real numbers with $b>a$, and $f$ be a continuous function mapping [a,b] to $\mathbb R$. Let $X = \{(x,f(x)) : x \in [a,b]\}$.
Now sweep out a planar area by rotating $X$ around the origin through some angle $\theta$ in $(0, 2 \pi]$. 
When $f(x) = 0$, this yields your (sectors of, when $\theta<2 \pi$) annuli (or disks, when $a=0$).
We can think of your case of parallelograms as being the limit when $a$ is very, very large compared to $b-a$; and where $f(x) = (x-a)c$; we're (very loosely speaking!) rotating the figure around "a point at infinity". (This corresponds to Hagen's set of translations).
But we aren't limited to a linear $f$ in either case; the only requirement is that there be no "self-intersections" as we sweep out the figure. And that can be satisfied if $d(x, f(x))$ (the distance from the point $(x, f(x))$ to the origin) is injective; i.e., for all $x_0, x_1 \in [a,b]$, $d(x_0, f(x_0)) = d(x_1, f(x_1)) \leftrightarrow x_0 = x_1$.
So for example: Take a triangle shaped wedge out of one end of your sector-of-an-annulus, and slap it onto the other end. Assuming the triangle isn't too acute, the resulting figure will have the property you describe. 
Lots of much more complicated possibilities can be constructed, as long as we obey the constraint of "distance from the origin is injective".
(Edit: replace "strictly increasing" with "injective".)
