 Solution

Let $B_c^{ab}$ denote the split-fuse diagram with a charge-$c$ edge joining the lower fusion vertex to the upper splitting vertex, and outer strands $a,b$. Completeness in the isotopy normalization of anyon diagrams readsResolve the two strands inside the encircling normalized Kirby color loop in this way. Move that loop onto the intermediate charge-$c$ edge. The Kirby killing property multiplies it by $\mathcal{D}{\delta }_{c1}$, leavingThe vacuum anyon channel is allowed only for $b=\bar{a}$, and then $d_b=d_a$ and $d_1=1$. Erasing the vacuum edge turns $B_1^{a\bar{a}}$ into the cup-cap diagram $E_a$, which connects the incoming pair by a cap and the outgoing pair by a cup. Thus the requested two-strand Kirby projector isHere $P_1^{a\bar{a}}=E_a/d_a$ is the orthogonal projection onto total vacuum anyon charge; for $b\ne \bar{a}$ it is zero. Both strands in the source point upwards, so the condition is $b=\bar{a}$, not $b=a$ unless $a$ is self-dual.

The coefficient can be checked without a convention guess: stacking two cup-cap diagrams creates one charge-$a$ loop, so $E_a^2=d_a{E}_a$. Consequently $(E_a/d_a{)}^2=E_a/d_a$, as required of an orthogonal projection. Omitting $1/d_a$ would fail this check.