There are plenty of $(C,P)$ such that all sections through $P$
have the same area. The construction readily adapts to higher moments
about $C$ but not to perimeter. The existence of a single convex $P$
with two (or more) $P$ may be a very hard question; already in two dimensions
the "equichordal point problem" was open for 80 years and took an
Inventiones paper to solve [1].
Put $P$ at the origin,
and let $S$ be the unit sphere $\{v \in {\bf R}^3 : \|v\| = 1\}$.
For $v \in S$ let $f(v)$ be $\sup\{a \in {\bf R} : av \in C\}$.
The area of a plane section of $C$ is the integral of $\frac12 f(v)^2$
over the corresponding great circle on the unit sphere.
If $C$ is a sphere about the origin then $f(v)$ is constant.
But for the great-circle integrals of $\frac12 f(v)^2$ to be constant,
it suffices to have $f(v)^2 + f(-v)^2$ constant, which is easy to arrange:
let $g : S \to {\bf R}$ be any smooth (or even $C^2$) odd function,
and set $f(v) = (1 + \epsilon g)^{1/2}$ for some small $\epsilon > 0$.
The resulting body $C_\epsilon := \bigcup_{v \in S} [0,f(v)] v$
must be convex for $\epsilon$ small enough,
because $C_0$ is the unit sphere and the curvature form of $C_\epsilon$
varies continuously with $\epsilon$.
For a reasonably simple explicit example, take $g$ to be
the restriction to $S$ of one of the coordinates of ${\bf R}^3$,
which makes $C_\epsilon$ a solid of revolution for the curve
given in polar coordinates by $r = (1 + \epsilon \cos \theta)^{1/2}$.
If I did this right, the curve, and thus $C_\epsilon$,
is convex if and only if $|\epsilon| \leq 2/3$.
Conversely, in any solution $f(v)^2 + f(-v)^2$ must be constant.
This follows from known properties of the Funk-Radon integral transform
${\cal F}$ on functions on the unit sphere, where $({\cal F}F)(x)$
is the integral of $F$ over the great circle orthogonal to $x$.
Clearly ${\cal F}F$ is even, and ${\cal F}F$ is the zero function if
$F$ is odd (which is what we used); Funk showed for even functions
${\cal F}F$ determines $F$, so if ${\cal F}F$ is the zero function then
$F$ is odd. See the references cited in Guillaume Aubrun's answer.
The construction readily adapts to any dimension $d$:
make $f(v)^{d-1} + f(-v)^{d-1}$ constant.
For constant moments of inertia about $P$,
change the exponent $d-1$ to $d+1$.
Reference
[1] Marek R. Rychlik (1997):
"A complete solution to the equichordal point problem of
Fujiwara, Blaschke, Rothe and Weitzenböck",
Inventiones Mathematicae 129 (1), 141--212.
(From Wikipedia's "Equichordal point problem" entry)
