Opposite Clifford-Algebra for a symmetric bilinearform $\beta$ on a $\mathbb{K}$-vectorspace $V$ the associated Clifford Algebra $Cl(\beta)$ is the associative algebra with unit subject to the relations $$v\cdot v=\beta(v,v)\cdot 1\qquad\forall v\in V.$$It is then often said that $Cl(-\beta)$ is isomorphic to the opposite Algebra $Cl(\beta)^\text{op}$. Why is that?
Cheers, Robert
 A: First, care is needed! We need the right notion of "op", for behold
$Cl(1) \simeq \mathbb{R}^2 \not\simeq \mathbb{C} \simeq  Cl(-1)$, which as algebras are both commutative!
Over $\mathbb{R}$, the Clifford algebras arise naturally in the braided symmetric monoidal category of $\mathbb{Z}/2$-graded vector spaces, or "super-vector spaces" (alternatively...), as (super-)tensor products of the basic algebras $\mathbb{R}[\varepsilon]$ and $\mathbb{R}[i]$, where $\varepsilon$ and $i$ denote odd roots of $\pm 1$.
As with ordinary vector spaces, there is an isomorphism
$$ \tau : A \otimes B \to B \otimes A, $$
and you need something like this to make an algebra of $A\otimes B$, given the algebra structures on $A$ and $B$ separately:
$$ (A\otimes B) \otimes (A\otimes B) \overset{A\otimes \tau\otimes B}\longrightarrow (A\otimes A)\otimes(B\otimes B) \overset{m_A\otimes m_B}\longrightarrow A \otimes B .$$
Figuring out different ways to define $\tau$ can be confusing at first, but you'll get used to it.
Now, in ordinary vector spaces, the "opposite" product $m^{op}$ is derived from $m$ by pre-composition with the transpose on $A\otimes A$:
$$ m^{op} = A\otimes A \overset{\tau}\to A\otimes A \overset{m}\to A, $$
and the same expression is the "right way" to define the opposite multiplication on a superalgebra; only the $\tau$ means something slightly different for superspaces vs. for ordinary vector spaces.
So, the exercises to start with are to understand $\tau$, and then to verify that $m_{-1}^{op} = m_1 $ and vice-versa --- or, understanding $\tau$, to carry-out the calculation over clifford algebras in full generality.
My favourite reference for superalgebras is Trimble's notes, but now you've read the words "symmetric" and "braided" and "monoidal" together, you can go on digging. Have fun!
A: There are different conventions for Clifford algebras, which is confusing.
If I understand your question correctly.
If the  convention for the bilinear form is
$
v . v=\beta(v,v)\cdot 1\qquad\forall v\in V
$
then the algebra is denoted as $\mathcal{Cl}_{p,q}$.
In the opposite convention
$v . v= - \beta(v,v)\cdot 1\qquad\forall v\in V
$
it is equivalent to the algebra $\mathcal{Cl}_{q,p}$.
