Let $A, B \in \mathbb{R}^{n \times n}$, $N(B)$ let denote the nullity of $B$. Let $C$ and $N(B)$ are complementary subspaces of $\mathbb{R}^n$. Let $P_c$ denotes the oblique projector onto $C$ along $N(B)$. What is its physical meaning?
If $p_o$ is orthogonal projector onto $C$ then i know that $p_o(x)$ for any $x \in \mathbb{R}^n$ is that element of $C$ which is nearest to $N(B)$. Am I right? Please tell me the difference and correct me if I am wrong.
You have a few things wrong.
First, note that because $N(B)$ is the orthogonal complement to $C$, the "oblique projector onto $C$ along $N(B)$" is also just the orthogonal projection onto $C$. That is, if $p_0$ is the orthogonal projector onto $C$ and $P_C$ is the oblique projector onto $C$ along $N(B)$, then $p_0(x) = P_Cx$.
Because $p_0$ is the orthogonal projection onto $C$, $p_0(x)$ is the element of $C$ which is nearest to x, not to $N(B)$.
If want to use the definition of an oblique projection, then we can say that any $x$ can be uniquely expressed as $x = x_C + x_{N(B)}$, where $x_C \in C$ and $x_{N(B)} \in N(B)$. The function $p_(x)$ takes the input $x$ and produces the vector $x_C$.
Because $N(B) = C^\perp$, it turns out that $x_C$ is the nearest element of $C$ to $x$.
An example where the projectors are different: take $C = \{(t,0): t \in \Bbb R\}$. Then the orthogonal projection onto $C$ is the transformation $p_{\perp}(x,y) = (x,0)$.
On the other hand, we could take an oblique projection, such as the projection along $B = \{(t,-t) : t \in \Bbb R\}$. This projection is given by $p_B(x,y) = (x+y,0)$.
Notice that for $p_\perp$, we have $p_{\perp}(x,y) = 0$ exactly when $(x,y)$ is in the orthogonal complement of $C$, which is to say that $(x,y)$ is on the $y$-axis. On the other hand: for $p_B$, we have $p_B(x,y) = 0$ exactly when $(x,y)$ is an element of $B$.
Yet another way to look at it: consider what happens to the point $(2,-1)$ under both transformations. Because $(2,-1) = (2,0) + (0,-1)$ is a sum of an element of $C$ with an element of $C^\perp$, we have $p(2,-1) = (2,0)$. On the other hand, because $(2,-1) = (1,0) + (1,-1)$ is a sum of an element of $C$ with an element of $B$, we can write $p_B(2,-1) = (1,0)$.