= Solution
\b[Yes, for $p=2$.] The displayed <group presentation> is that of the infinite <dihedral group>. Put $u=at$ and $v=t$. Then $v^2=1$ and
$$
u^2=atat=a(tat^{-1})t^2=aa^{-1}=1.
$$
Conversely, from $u^2=v^2=1$, set $a=uv$ and $t=v$; then $tat^{-1}=vu=a^{-1}$. These inverse substitutions give
$$
G_2\cong\langle u,v\mid u^2,v^2\rangle\cong C_2*C_2.
$$
Each relator has free-group $2$-root exponent one, so
$$
\boxed{\operatorname{def}_2(\langle u,v\mid u^2,v^2\rangle)=2-\frac12-\frac12=1.}
$$
The change of presentation matters: the original presentation's mixed conjugation relator has weight one and would give a smaller <p-deficiency>.
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