1. Write the matrix and the given conditions

   Any matrix in $T_p$ is of the form $A=\begin{pmatrix} a & b \\ c & a \end{pmatrix}, \qquad a,b,c\in\{0,1,2,\dots,p-1\}.$

   Its trace is $\operatorname{tr}(A)=a+a=2a.$

   Its determinant is $\det(A)=a^2-bc.$

2. Condition on trace

   We need trace not divisible by $p$.

   Since $p$ is an odd prime, $p\nmid 2$. Therefore, $p\mid 2a \iff p\mid a.$

   Hence, $\operatorname{tr}(A) \text{ is not divisible by } p \iff a\not\equiv 0 \pmod p.$

   Since $a\in\{0,1,\dots,p-1\}$, this means $a\in\{1,2,\dots,p-1\}.$ So there are exactly $p-1$ choices for $a$.

3. Condition on determinant

   We need $p\mid \det(A)=a^2-bc,$ i.e. $bc\equiv a^2 \pmod p.$

   Now fix some nonzero $a$.

   Since $a\not\equiv 0\pmod p$, we have $a^2\not\equiv 0\pmod p$. Thus the congruence $bc\equiv a^2 \pmod p$ implies that both $b$ and $c$ must be nonzero modulo $p$.

4. Count $(b,c)$ for a fixed nonzero $a$

   For each choice of nonzero $b\in\{1,2,\dots,p-1\}$, there is a unique $c$ modulo $p$ given by $c\equiv a^2 b^{-1} \pmod p,$ because every nonzero element modulo $p$ has a unique inverse.

   Therefore, for each fixed nonzero $a$, the number of pairs $(b,c)$ satisfying $bc\equiv a^2\pmod p$ is exactly $p-1.$

5. Total count

   Number of choices for $a$: $p-1.$

   For each such $a$, number of valid $(b,c)$ pairs: $p-1.$

   Hence total number of matrices is $(p-1)(p-1)=(p-1)^2.$

6. Match with options

   $(p-1)^2$ corresponds to Option C.

7. Compare with stored correct answer

   Stored correct answer is C, which matches our result.

Final Answer: Option C, $\boxed{(p-1)^2}$