Theorem: Suppose the $n \times n$ matrix $A$ has $n$ linearly independent eigenvectors. If these eigenvectors are the columns of a matrix $S,$ then $S^{-1}AS$ is a diagonal matrix $\Lambda$. The eigenvalues of $A$ are on the diagonal of $\Lambda$.
$S^{-1}AS = \Lambda$ (A diagonal Matrix with diagonal values representing eigen values of A)$ = \begin{bmatrix} \lambda _{1} & & & & \\ & \lambda _{2} & & & \\ & & . & & \\ & & & . & \lambda _{n}\\ & & & & \end{bmatrix}$
Now if $A$ is diagonalizable, $S^{-1}$ must exist. What is $S?$ $S$ is a matrix whose columns are eigen-vectors of matrix $A.$
Now $S^{-1}$ would only exist if $S$ is invertible. And $S$ would be invertible if all rows and columns of $S$ are independent.
If we would have same eigen-vectors, then $S^{-1}$ won't exist and hence $A$ won't be diagonalizable.
Even if a matrix $A$ has same eigen values, it does not mean that it is not diagonizable. Take the trivial example of the Identity Matrix $I.$
$\begin{bmatrix} 1 & & \\ & 1& \\ & & 1 \end{bmatrix}$
The Eigen values are $1,1,1$ and Eigen vectors are $\begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}$ , $\begin{bmatrix} 0\\ 1\\ 0\end{bmatrix}$, $\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}$
We form $S=\begin{bmatrix} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 & 0& 1 \end{bmatrix}$ and this is invertible.
So, this makes $S^{-1} I S$ as $I =\Lambda$ which is in-fact a diagonal matrix.
So, even if we have same eigenvalues the matrix may or may not be diagonalizable. But yes, we need full $n$ set of linearly independent eigenvectors for this matrix $A$ of size $n \times n$ to be diagonalizable.
Now our problem says that we have a matrix $P$ whose only eigenvectors are multiples of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$.
Means we have only $\begin{bmatrix} 1\\ 4 \end{bmatrix}$ as independent eigen vector. Surely, this matrix is not diagonalizable.
Since, eigen vectors are multiples of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$, we have repeated eigen values.
Let us assume $\lambda _{1}$ and $\lambda _{2}$ are different eigen values.
Since, we have only eigen vectors multiple of $\begin{bmatrix} 1\\ 4 \end{bmatrix}$.
Let this vector be $x_1.$
So, $Px_1=\lambda _{1}x_1$
$\implies Px_1=\lambda _{2}x_1$
$\implies \lambda _{1}x_1=\lambda _{2}x_1$
Multiply by $x_1-1$ we get $\lambda _{1} =\lambda _{2}.$ Means, same Eigen vector gives same Eigen Value.
So yes, $P$ has repeated Eigen Values. (II) statement is true.
Now Statement (I) is not true. We cannot say this statement with exact surety.
Consider a matrix $A=\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$
It's eigenvales are $\lambda _{1}=\lambda _{2}=0$.
All the eigenvectors of this $A$ are multiples of vector $(1,0)$
$\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}. x=\begin{bmatrix} 0\\ 0 \end{bmatrix}$
or $x=\begin{bmatrix} c\\ 0 \end{bmatrix}$.
This matrix surely is not diagonalizable but this matrix $A$ has determinant $= -1.$
Since determinant is not $0,$ so $A^{-1}$ exists.
Hence, (II) and (III) are surely valid under all cases of this question.
For (I) part, Matrix $P$ will not have an inverse when $det(P)=0$ and this implies one of the eigenvalue of $P$ is zero. But in question, no where it is mentioned about what are the eigen values. So, I is not necessarily true.
Answer-(D) (Remember options says "necessarily true")