We say that cartesian equation of a plane is $ax+by+cz=d--(1)$ , but lets say P is a mobile point (x,y,z), b is a fixed point that lies on a plane and $n$(a,b,c) is normal to the plane hence should be normal to b as well.
To derive the equation we use dot product $$p.n=b.n$$ $$or, xa+yb+zc=b.n$$ $$or,ax+yb+cz=b.n---(2)$$ Now if we look at equation $(1)$ and $(2)$.As b and n are perpendicular to each other by definition,won't their dot product $b.n=0$ ? $$so, ax+yb+cz=0$$ I might have missed something really important out here which is confusing me please help.
A normal vector to a given plane is perpendicular to the difference of two vectors whose terminal points (with initial points at the origin) are on the plane.
The points $b$ of $\mathbb{R}^3$ such that $\vec{b}\cdot\vec{n}=0$ are precisely the points of the plane $ax+by+cz=0$.
Hence if $d\ne 0$, a normal vector $\vec{n}$ will not be perpendicular to a vector whose terminal point (with initial point at the origin) is on the plane.
Thus, in the context of your question, the dot product $\vec{b}\cdot\vec{n}$ will not be equal to zero (unless $d=0$).