So, I want to expand this term but facing difficulty with the vector and tensor product. The term is :
$\textbf{p}$ is a vector; and $\mathbb{K}$ is a tensor
$\nabla\cdot[\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})]$
There is divergence of all terms and when expanding this i am stuck with how to deal with tensor and vector product.
Thank you for helping me out.
If
when you write $$\nabla\textbf{p}$$ you mean the 2-tensor that results from taking the "gradient" (covariant derivative) of $\mathbb{p}$ (which is, actually kind of weird),
when you write $$\mathbb{K} \cdot \nabla\textbf{p}$$ you mean the scalar that results from taking the "interior product" of (contracting) $\mathbb{K}$ with $\nabla \mathbf{p}$,
when you write $$\nabla\cdot[\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})]$$ you mean the scalar that results from taking the divergence of $\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})$ (which is a vector, since it is the product of a vector with a scalar)
then you can use the identity $$\nabla\cdot(f\mathbf{v}) = \nabla f\cdot\mathbf{v} + f(\nabla\cdot\mathbf{v})$$ to get $$\nabla\cdot[\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})] = (\nabla\cdot\mathbf{p})(\mathbb{K}\cdot\nabla\textbf{p}) + \mathbf{p} \cdot \nabla(\mathbb{K}\cdot\nabla\textbf{p})$$
If want to expand it further I would recommend you to rewrite your expression in abstract index notation or something similar
$$\begin{align} \nabla\cdot[\textbf{p}(\mathbb{K}\cdot\nabla\textbf{p})] &= \nabla_a(p^a{K^b}_c\nabla_bp^c)\\ &= \nabla_a p^a{K^b}_c\nabla_bp^c + p^{a}\nabla_a({K^b}_c\nabla_bp^c)\\ &=\nabla_a p^a{K^b}_c\nabla_bp^c + p^{a}[\nabla_a{K^b}_c\nabla_bp^c + {K^b}_c\nabla_a\nabla_bp^c] \end{align}$$
but that doesn't give us anything that can be written in vector calculus notation. (Well, even that $\nabla\textbf{p}$ is not common).