= Solution
The <normal closure> $\langle\!\langle B\rangle\!\rangle_G$ is the smallest <normal subgroup> of $G$ containing $B$, equivalently the intersection of all normal subgroups containing $B$. Any such <normal subgroup> contains every $gb^{\pm1}g^{-1}$ and hence every finite product of these elements. Thus $N\subseteq\langle\!\langle B\rangle\!\rangle_G$.
Include the empty product so that $1\in N$, in particular when $B=\varnothing$. Products of elements of $N$ are obtained by concatenating their factor lists. Inversion reverses the list and replaces $\epsilon_i$ by $-\epsilon_i$, so $N$ is closed under inverses. For any $k\in G$,
$$
k\left(\prod_i g_i b_i^{\epsilon_i}g_i^{-1}\right)k^{-1}
=\prod_i(kg_i)b_i^{\epsilon_i}(kg_i)^{-1}\in N.
$$
Therefore $N$ is a <normal subgroup> and it contains $B$. Minimality of the <normal closure> gives the reverse inclusion, proving
$$
\boxed{N=\langle\!\langle B\rangle\!\rangle_G.}
$$
The notation $n\in\mathbb N$ in a finite-product description is understood to allow the empty product; otherwise the empty generating subset would require a separate identity convention.
In a <group presentation>, $X$ is a set of formal generators and $R$ is a subset of the <free group> $F(X)$, consisting of relators which are required to become the identity. The presented <group> is
$$
\boxed{\langle X\mid R\rangle=F(X)/\langle\!\langle R\rangle\!\rangle_{F(X)}.}
$$
The <universal property of a group presentation> says that a function $X\to H$ extends to a <group homomorphism> from the presented <group> precisely when every <relator> evaluates to $1$ in $H$: first extend freely to $F(X)$, then factor through the quotient because its <kernel> contains the <normal closure> of $R$.
A <rotation> $\alpha$ by $2\pi/n$ and a <reflection> $\beta$ of a regular $n$-gon satisfy
$$
\alpha^n=1,\qquad\beta^2=1,\qquad\beta\alpha\beta=\alpha^{-1}.
$$
The last identity holds because <reflection> reverses the sense of <rotation>, and it implies $\alpha\beta\alpha\beta=1$. Hence all defining relators vanish, giving a homomorphism $\theta:D_{2n}\to M_n$ with $\theta(\sigma)=\alpha$ and $\theta(\tau)=\beta$.
To prove it is an <isomorphism> without presupposing the order of the presented <group>, use $\sigma\tau=\tau\sigma^{-1}$. Move every occurrence of $\tau$ to the left, cancel pairs by $\tau^2=1$, and reduce exponents modulo $n$. Every element has one of the forms
$$
\sigma^j\quad\text{or}\quad\tau\sigma^j,\qquad0\leq j<n.
$$
Thus $|D_{2n}|\leq2n$. For the usual <regular polygon> with $n\geq3$, the $n$ rotations $\alpha^j$ are distinct and the $n$ reflections $\beta\alpha^j$ are distinct; no <rotation> is a <reflection>, since their planar <orientation> behavior differs. The supplied description says these exhaust $M_n$, so $|M_n|=2n$, and every one lies in the image of $\theta$. Surjectivity and the order bound force $|D_{2n}|=2n$ and trivial <kernel>. Thus \b[$\theta$ is an <isomorphism>].
For the requested <Cayley graph>, use right multiplication by the generators. The $\sigma$ edges are the two directed 3-cycles
$$
1\to\sigma\to\sigma^2\to1,
\qquad
\tau\to\tau\sigma\to\tau\sigma^2\to\tau.
$$
The $\tau$ edges are the undirected pairs
$$
1\leftrightarrow\tau,\qquad
\sigma\leftrightarrow\tau\sigma^2,\qquad
\sigma^2\leftrightarrow\tau\sigma.
$$
These pairs use $\sigma^j\tau=\tau\sigma^{-j}$. Thus the underlying <graph> is a <triangular prism>, with the two <rotation> cycles oriented oppositely when <reflection> pairs are drawn as radial edges.
\Image[/past-exam-of-the-topology-and-groups-course-of-the-university-of-oxford/2010-dihedral-cayley.png]
{title=The Cayley graph of the dihedral group of order six}
{description=Blue arrows are right multiplication by sigma. Orange undirected edges are right multiplication by the involution tau.}
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