If we consider $P=\left \{ {C_1},{C_2},{C_3},{C_4},{C_5}\right \}$ and $X=\begin{bmatrix} x_1 &x_2 & x_3 & x_4 & x_5 \end{bmatrix}^T$
Since $P X = 0$
Then, $x_1C_1+x_2C_2+x_3C_3+x_4C_4+x_5C_5=0$
$\left ( Given, x_1=2k, x_2=5k, x_3=4k,x_4=3k,x_5=k \right )$
So, $2C_1+5C_2+4C_3+3C_4+C_5=0$
$C5=-2C_1-5C_2-4C_3-3C_4$
This proves that one column vector is dependent on the other $4$ independent column vectors.
Rank = number of linearly independent column vectors in a matrix which is $4$ here.
Other methods to solve this problem:
The dimension of the null space of matrix P is called the nullity of P.
Let say, $u=\begin{bmatrix} 2 &5 &4 & 3 & 1 \end{bmatrix}^T$
Since every other solution can be expressed in the form of given vector, so we can write $X=k\cdot u$. This means u form the basis of null$(P)$ and null space has dimension 1. Hence nullity$(P)=1$,
Or,
nullity is the no of free variables in the solution of $PX=0$.
Here, $X=\begin{bmatrix} 2k \\ 5k \\ 4k \\ 3k \\k \end{bmatrix}$
Clearly, $x_1,x_2,x_3,x_4$ are dependent on $x_5$. Only 1 free variable in solution.
so nullity$(P)$ =1.
By rank-nullity theorem,
rank$(P)+$ nullity$(P)=$ no of column of P
$r+1=5 \rightarrow r=4$
(Correct me if something wrong).