Let $A\in M_3(\mathbb R)$ which is not a diagonal matrix.Let $p$ be a polynomial in one variable with real coefficients & of degree 3 such that $p(A)=0$.
Can we conclude that $p=cp_A $ where $c\in \mathbb R $ and $p_A$ is acharacteristic polynomial of $A$?If not what kind of roots(real/complex) p should have in order to make such a conclusion.
Take $A= \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$ Then $p_A(x) = (x-1)^2 (x+1)$. Let $p(x) = (x-1) (x+1)(x+2)$, then $p(A) = 0$.
The condition is true iff the minimal polynomial and the characteristic polynomials are the same iff the dimension of each eigenspace of $A$ is one.
I found the latter condition at https://math.stackexchange.com/a/81473/27978.
No, one cannot conclude this. You cannot expect to ever conclude from $P[A]=0$ that $P$ is a scalar multiple of any given polynomial (like $P_A$), as any polynomial multiple $PQ$ of $P$ will also satisfy $(PQ)[A]=0$.
For instance for $A=0$ one has $P_A=X^3$ but any polynomial without constant term satisfies $p[A]=0$. Here a polynomial must have $0$ as at least triple root to be divisible (in the polynomial sense) by $P_A$. Since I missed that the question demands a non-diagonal matrix, I must modify a bit to make a "valid" example. I find that condition stupid because it is not preserved under change of basis, and there is no way in which such a condition could really help to prove anything (by contrast "not diagonalisable" could be relevant); however it does imply "not a multiple of the identity", so base change alone will not help to make my example valid.
First change the example by putting $1$ in the first entry, giving a diagonal matrix $A$ with diagonal $(1,0,0)$. Now $P_A=(X-1)X^2$ but $P[A]=0$ only forces $P$ to have roots $0,1$, so $P=X(X-1)$ (or any multiple $X(X-1)(X-a)$ of it of degree$~3$, another condition I overlooked) will satisfy $P[A]=0$. Now to make the example non-diagonal change to any basis not consisting entirely of eigenvectors. For instance $(1,1,0)$ is not an eigenvector, so replacing the first basis vector by it, we find that for $$ P=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix} \quad\text{one has}\quad P^{-1}AP=\begin{pmatrix}1&0&0\\-1&0&0\\0&0&0\end{pmatrix} $$ which has the same properties as $A$ (same characteristic polynomial, same condition $P[A]=0$) without being diagonal.
In case you are curious, the answer is still "no we cannot conclude" if one replace "non diagonal" by "non diagonalisable". Any matrix$~A$ with a single non-zero coefficient that is off-diagonal satisfies $A^2=0$, so $P[A]=0$ for every polynomial with at least a double root$~0$; however $P_A=X^3$ for such matrices.
In general there is a unique monic polynomial $\mu_A$ such that $P[A]=0$ is equivalent to $P$ being a polynomial multiple of $\mu_A$; it is called the minimal polynomial of$~A$. For the last example $A$ one has $\mu_A=X(X-1)$. For many matrices $A$ one has $\mu_A=P_A$, for instance this is the case whenever when $P_A$ has no repeated factors; the precise condition is a bit subtle to formulate.
As for the final question, for a degree three real polynomial, having some non-real root forces having three distinct roots in$~\Bbb C$ (since the complex conjugate of that first root is also a root, and a final root is real), but I find this a really awkward way to formulate a condition (no repeated factors is much better).
