Let us give an example that makes this as simple as possible.
Suppose your samples are all taken from a Bernoulli distribution, i.e. these are ``binary samples'', these are just $1$'s and $0$'s. Let those samples by denoted by: $x_1,x_2,...,x_n$. Here each $x_k$ is equal to either $1$ or $0$. Let $\mathbf{x}$ denote the entire vector of all those samples, so $\mathbf{x} = (x_1,x_2,...,x_n)$.
Now, since these samples are coming from a Bernoulli distribution it means there is an unknown parameter $c$ which represents the ``success rate''. The number/parameter $c$ is unknown, and it is supposed to represent how often the samples display $1$. Therefore, if $c=.9$ then we expect to see a lot of $1$'s in the vector $\mathbf{x}$ and if $c=.1$ then we expect to see mostly zeros instead.
Let us say, for example, that our sample vector $\mathbf{x}$ consists of $70$ observations of "$1$" and $30$ observations of "$0$". It is reasonable to guess that $c = \frac{70}{100} = .7$. However, because of random flucations in the data it could happen that the true value of $c=.65$ and the data was just more lucky and generated a bit more $1$'s.
The "posterior distribution", which you denote as $P(c|\mathbf{x})$ is supposed to quantify your uncertainty about the value of the $c$ parameter. In the example we are using, where $\mathbf{x}$ has 70 successes and 30 failures, it can be shown (perhaps, you can show this yourself!), that the posterior distribution looks like this,

From this picture you can see that the most reasonable choice for $c$ is $0.7$, but it could also be $0.8$ but much less likely, however when we reach $0.9$ it becomes very unreasonable. Instead of saying "more reasonable", or "less reasonable", ect, we make the language precise so there is no confusion about what we mean, and the posterior distribution is what quantities your uncertainty and likelihood of the unknown parameter $c$.