So for the first place I got stuck, showing that each root appears $n/d$ times, it seems that, if you let $H'$ be the subgroup that fixes $F(\alpha)$, then since there are $n/d$ cosets of $H$ in each coset of $H'$, and since each coset of $H'$ fixes a root, each root appears $n/d$ times, once for each coset of $H$.
For the matrix thing, if $k_1,\ldots,k_{n/d}$ is a basis of $K$ over $F(\alpha)$, then since $1, \alpha,\ldots, \alpha^{d-1}$ is a basis for $F(\alpha)$ over $F$, a basis for $K$ over $F$ is given by the products $k_i\alpha^j$, where $1\leq k\leq n/d$ and $0\leq j\leq d-1$. Then, the matrix of "multiplication by $\alpha$" with respect to this basis is the direct sum of $n/d$ copies of the $d\times d$ companion matrix for the minimal polynomial of $\alpha$, from which the trace and determinant are easily seen to be as desired.
Let's start off by looking at the characteristic polynomial of the $2 \times 2$ matrix
$A = \begin{bmatrix} 0 & 1 \\ -a & -b \end{bmatrix}: \tag 1$
$\det (A - \lambda I) = \det \left (\begin{bmatrix} -\lambda & 1 \\ -a & -b - \lambda \end{bmatrix} \right ) = \lambda^2 + b\lambda + a; \tag 2$
we see from (2) that we may always present a $2 \times 2$ matrix with given characteristic polynomial $\lambda^2 + b\lambda + a$ in the form $A$; for example, if he quadratic is $\lambda^2 + 5\lambda + 1$, as in the present problem (tho' I have replaced $p$ with $\lambda$), we may take
$P = \begin{bmatrix} 0 & 1 \\ -1 & -5 \end{bmatrix}, \tag 3$
which may be easily checked:
$\det(P - \lambda I) = -\lambda(-5 - \lambda) - 1 ( -1) = \lambda^2 + 5\lambda + 1. \tag 4$
The matrices $A$ and $P$ above are part of a general paradigm for constructing matrices with a given characteristic polynomial, and it extends to higher dimensions. Now, there are a very many matrices possessed of a given characteristic polynomial, since it is a similarity invariant; that is, the characteristic polynomials of $X$ and $S^{-1}XS$ are always the same; thus it behooves us to find a matrix of particularly simple, general form for a given polynomial. If
$q(\lambda) = \displaystyle \sum_1^n q_i \lambda^i, \; q_n = 1, \tag5$
we define $C(q(\lambda))$ to be the $n \times n$ matrix
$C(q(\lambda)) = \begin{bmatrix} 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ldots & 0 \\
\vdots & \vdots & \ldots & \vdots & \vdots \\ -q_0 & -q_1 & -q_2 & \ldots & -q_{n - 1} \end{bmatrix}; \tag 6$
that is, $C(q(\lambda))$ has all $1$s on the superdiagonal, the negatives of the coefficients of $q(\lambda)$ on the $n$-th row, and $0$s everywhere else. We have
$C(q(\lambda) - \lambda I = \begin{bmatrix} -\lambda & 1 & 0 & \ldots & 0 \\ 0 & -\lambda & 1 & \ldots & 0 \\
\vdots & \vdots & \ldots & \vdots & \vdots \\ -q_0 & -q_1 & -q_2 & \ldots & -q_{n - 1} - \lambda \end{bmatrix}; \tag 7$
it is easy to see, by expanding in minors along the $n$-th row, that
$\det(C(q(\lambda)) - \lambda I) = q(\lambda); \tag 8$
also, since a matrix and its transpose have equal determinants, the transposed form $C(q(\lambda))$, $C^T(q(\lambda))$, also gives rise to the same characteristic polynomial. The matrices $C(q(\lambda))$ and $C^T(q(\lambda))$ are known as companion matrices for the polynomial $q(\lambda)$; the linked article has more of the story.
Best Answer
It is just a misconception of sign i.e. $(-1)^3\{\lambda^{3}-(Trace(A))\lambda^{2}+\lambda-|A|\}$ = $|\lambda I - A| $. So your last option is correct.