Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 4 / 5 / ii / Solution

Yes, for $p=2$. The displayed group presentation is that of the infinite dihedral group. Put $u=at$ and $v=t$. Then $v^2=1$ andConversely, from $u^2=v^2=1$, set $a=uv$ and $t=v$; then $ta{t}^{-1}=vu=a^{-1}$. These inverse substitutions giveEach relator has free-group $2$-root exponent one, soThe change of presentation matters: the original presentation's mixed conjugation relator has weight one and would give a smaller p-deficiency.