We have an odd prime $p$. Consider the set

$ T_p \;=\; \Bigl\{ A \;=\;\begin{pmatrix} a & b \\[6pt] c & a \end{pmatrix} \;:\; a,b,c \in \{0,1,2,\dots,p-1\} \Bigr\}. $

We want to count the number of such matrices $A$ that satisfy:

$\text{trace}(A) \not\equiv 0 \pmod{p}$.

Since $\text{trace}(A) = a + a = 2a$, this means

$ 2a \;\not\equiv\; 0 \pmod{p}. $

Because $p$ is an odd prime, $2$ is invertible $\bmod\,p$.

Hence $2a \equiv 0 \pmod{p}$ if and only if $a \equiv 0 \pmod{p}$.

Thus the condition

$\text{trace}(A) \not\equiv 0 \pmod{p}$

is equivalent to

$ a \;\neq\; 0 \pmod{p}. $

In other words, $a$ must be a nonzero element modulo $p$; so

$a\in\{1,2,\dots,p-1\}.$

$\det(A) \equiv 0 \pmod{p}$.

For the matrix

$\begin{pmatrix} a & b \\ c & a \end{pmatrix},$ we have

$ \det(A) \;=\; a^2 - bc. $

The condition $\det(A)\equiv 0\pmod{p}$ means

$ a^2 \;-\; bc \;\equiv\; 0 \pmod{p} \quad\Longleftrightarrow\quad bc \;\equiv\; a^2 \pmod{p}. $

Putting these two pieces together:

$a$ is nonzero ($a \in \{1,\dots,p-1\}$).

$bc \equiv a^2 \pmod{p}.$

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Counting the solutions

We must count the number of triples $(a,b,c)$ with $a\neq 0$ and $bc \equiv a^2$.

Range for $a$

Since $a\in \{1,2,\dots,p-1\}$, there are $p-1$ possible values of $a$.

Condition $bc \equiv a^2$ for given $a$

Since $a \neq 0$, $a^2$ is also nonzero $\bmod\,p$.

If $b = 0$, then $bc \equiv 0$, which would only be equal to $a^2$ if $a^2 \equiv 0$. But $a^2\neq 0$ since $a\neq 0$. So $b=0$ gives no solutions.

Similarly, if $c=0$, then $bc=0$, which again cannot equal the nonzero $a^2$. So $c=0$ gives no solutions either.

Therefore $b$ and $c$ must both be nonzero modulo $p$.

Once $b$ is chosen to be any nonzero element in $\{1,\dots,p-1\}$, there is a unique $c \equiv a^2 \,b^{-1} \pmod{p}$. Hence:

For each nonzero $b$ (that is, $b\in\{1,\dots,p-1\}$), there is exactly one $c\in\{1,\dots,p-1\}$ satisfying $bc \equiv a^2$.

Consequently, for each fixed nonzero $a$, the number of $(b,c)$-pairs is $p-1$.

Putting it all together

We have $p-1$ choices for $a\neq 0$.

For each such $a$, there are $p-1$ pairs $(b,c)$ satisfying $bc\equiv a^2$.

Therefore, in total, the number of matrices is

$ (p-1)\,\times\,(p-1) \;=\; (p-1)^2. $

Hence the correct answer matches option C, which is

$ \boxed{(p-1)^2.} $