Parity of proton is 1? I have found from Wikipedia that 
"a parity transformation is the flip in the sign of spatial coordinates".
Now when we operate parity operator, does that mean we are taking any physical entity at ${\bf x}$ to $-{\bf x}$. Or we are just reverting axes of the co-ordinate system?
However if we take parity transformation as active transformation then what does it mean that parity of proton is 1? Doesn't it anyway depend on the origin of the co-ordinate system? 
Please elucidate the meaning of the parity. I get too much confused whenever I hear  "parity"!
 A: 
Now when we operate parity operator, does that mean we are taking any physical entity at x to −x. Or we are just reverting axes of the co-ordinate system?

Well, either operation should adhere to the same rules, and you mention the correct term: it depends on whether we see the operation as active or passive. Either view has the same end result: we move "$\vec{x}$ to $-\vec{x}$", as you say.
It does not depend on the origin, since we change the coordinate system, not just in the initial state, but in the entire operation. Feynman described it as:

So if the laws of physics are symmetrical, we should find that if some demon were to sneak into all the physics laboratories and replace the word “right” for “left” in every book in which “right-hand rules” are given, and instead we were to use all “left-hand rules,” uniformly, then it should make no difference whatever in the physical laws.

Feynman lectures in physics, chapter 52, section 5

There is a "namespace collision" at play when one talks about parity in physics, which can be a source of initial confusion (at least it was for me): both the symmetry and the conserved quantity are often called just "parity". In contrast, we use different terms for e.g. the symmetry "time" and the corresponding conserved quantity "energy". The parity symmetry is sometimes called "inversion", which avoids this problem. For more information on symmetries and respective currents and charges, see Noether's theorem (note that parity is a discrete and not continuous symmetry, though).
When we say that "the parity of a proton is $+1$" (or just "$+$", or "even"), we are talking about a quantity inherent to the particle in itself — a "parity charge", if you will — that is found to be multiplicatively conserved (including spatial parity) in certain interactions. Every elementary particle is given such an intrinsic parity — the difference between "intrinsic" and "spatial" parity seems to be at the root of your question. I will later show some examples of the interplay between these concepts and how they affect decays.
The parity operation is found to be a symmetry (at current knowledge) for gravity (though that is mostly irrelevant for physics at this scale), the strong force and electromagnetism, but not for the weak force. When fundamental physics says that "parity is a conserved quantum number", it implies strong and EM interactions only, since we experimentally know it not to be true for the weak interaction.
The intrinsic parity $\pi$ of the proton is by convention set to $+1$. All "regular" fermions (half-integral spin) also have $\pi=+1$, and their antiparticles have opposite parity. Bosons (integral spin) and their antiparticles have the same parity. These intrinsic quantities are the ones that should be (multiplicatively) conserved under strong and EM interactions, together with the spatial parities — most commonly the relative orbital angular momentum $l$, which has parity $(-1)^l$.
To reiterate what is perhaps the main point: the proton parity is defined to be $+1$. It can be shown that the intrinsic parity must be either of $\pm 1$: heuristically, we can look at the measurable quantity of a radial wave function $\Psi(\vec{r})$, which is $|\Psi(\vec{r})|^2$. If parity is a true symmetry, then we know that $$|\Psi(\vec{r})|^2 = |\Psi(-\vec{r})|^2$$ which implies that $\Psi(\vec{r})=\pm\Psi(-\vec{r})$, i.e. the parity operation $\pi$ applied on $\Psi$ is either $+1$: $\pi\Psi(\vec{r}) = +\Psi(-\vec{r})$, or $-1$: $\pi\Psi(\vec{r}) = -\Psi(-\vec{r})$. Alternatively, we can say that since parity applied twice gives us back the original system, then $\pi^2=1$ which implies $\pi = \pm 1$. This is not at all rigorous — the statement should rather be: "any non-degenerate energy state has either of $\pi=\pm 1$" (see the final link to Feynman's lectures for a derivation).
Why can we "define" the intrinsic parity for the proton arbitrarily? Well, we can measure the relative parities among the fundamental particles (i.e. that an antiproton has opposite parity of a proton) through the conservation laws, but the "absolute phase" of the parity operator holds no physical significance, since the observable quantity is $|\Psi|^2$. It is a bit like how we define the electron to have negative electric charge, and not the other way around.

Example 1: The $\eta$ meson ($\pi=-1$) is observed to decay strongly to three pions: $$\eta → \pi⁺\pi⁻\pi⁰$$ (sadly another nomenclature collision with $\pi$ here…). The intrinsic parity of these are all $-1$, so the ending parity is $(-1)^{3+l}=-1$, if we have $l=0$ among the products. By reasoning on available mass only and other conserved quantum numbers, we should be able to see $\eta → \pi⁺\pi⁻$ as well, but what happens to parity? The end products would have parity $(-1)^{2+l}$, but we must also conserve the angular momentum, so we are not free to choose $l$! In fact, $\eta$ has spin $0$, as do $\pi^\pm$, so we would have to have $l=0$, and thus $\pi=1$ in the end products — this would not conserve parity, and is indeed not seen in nature.
Example 2: in Introductory nuclear physics by Krane, one should determine the intrinsic parity of the $\phi$ meson by observing the strong decay $$\phi → K⁺K⁻$$ given that $\phi$ has spin $1$ and $K^\pm$ have spin $0$. To conserve angular momentum, the relative orbital angular momentum among the end products must thus correspond to the spin of $\phi$, which was $1$. Particles and antiparticles among bosons have equal parity, so without knowing the intrinsic parity of $K^\pm$, their product must be $+1$, so the final parity will be $(+1)(-1)^l = -1$ — since the strong interaction preserves parity, the $\phi$ must therefore also have parity $-1$.
Example 3: However, the decay $$K⁺ → \pi⁺ \pi⁰$$ was experimentally seen to occur at a relatively high rate (${\sim}21\%$). By the same reasoning as in example 1, this would be forbidden strongly by parity conservation (alternatively, if we draw a quark scheme we also see that it violates conservation of strangeness, which is also strongly forbidden). Then how can it happen? Because the weak force does not obey the parity symmetry! This is a weak decay (which also permits strangeness changing). It is not "drowned" by otherwise dominating strong decays since there are no strangeness conserving channels energetically available, and thus the weak interaction is a necessary component in the decay process.

That became a lot of text, but still leaves a lot unsaid; perhaps even part of the main question. Feynman has another explication of the parity symmetry in Feynman lectures on physics, chapter 17, section 2. I am of course also open to corrections — that is one of the main motivations of writing in the SE network.
