$M = \pmatrix{1 + p& 0& 0& p\\p& 2p& 1 + 3p& p\\2p& 0& 0& 0\\0& 4p& 0& 2p\\p& 2p& 3p& p}$ has rank $4$ for for a variable $p\in\mathbb{C} \text{ or }\mathbb{R}$, does this mean: for any complex number $p$ the columns are linearly independent set of vectors in $\mathbb{C}^5 \text { or }\mathbb{R}^5$?
And, regarding the assurance of its rank $4$, I put it in Matlab and asked rank($M$), and it gives result $4$
For $p=0$, the last column is equal to zero, so in this case rank is clearly at most $3$.
Let us have a look on the case $p\ne0$. A rather standard method to find matrix rank is to use elementary row operations and/or elementary column operations. Notice that since we assume that $p\ne0$, we are allowed to divide by $p$.
$\begin{pmatrix} 1+p& 0 & 0 & p\\ p &2p &1+3p& p\\ 2p & 0 & 0 & 0\\ 0 &4p & 0 &2p\\ p &2p &3p & p \end{pmatrix}\sim \begin{pmatrix} 1+p& 0 & 0 & p\\ p &2p &1+3p& p\\ 2 & 0 & 0 & 0\\ 0 & 4 & 0 & 2\\ 1 & 2 & 3 & 1 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 & 0\\ 1+p& 0 & 0 & p\\ p &2p &1+3p& p\\ 0 & 4 & 0 & 2\\ 1 & 2 & 3 & 1 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & p\\ 0 &2p &1+3p& p\\ 0 & 2 & 0 & 1\\ 0 & 2 & 3 & 1 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 &1+3p& 0\\ 0 & 2 & 0 & 1\\ 0 & 0 & 3 & 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 1\\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 &1+3p& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 &1+3p& 0 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0& 0 \end{pmatrix}$
We see that the rank (in the case $p\ne0$) is indeed equal to $4$.
And to answer your additional question: Yes, this means that for $p\ne0$ the columns are linearly independent.