= Solution
Let $B^{ab}_c$ 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> reads
$$
1_{a\otimes b}=\sum_c\sqrt{\frac{d_c}{d_ad_b}}\,B^{ab}_c.
$$
Resolve 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 $\mathcal D\delta_{c1}$, leaving
$$
K_\Omega(a,b)=\mathcal D\sqrt{\frac{d_1}{d_ad_b}}\,B^{ab}_1.
$$
The <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 $B^{a\bar a}_1$ 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> is
$$
\boxed{K_\Omega(a,b)=\delta_{b,\bar a}\frac{\mathcal D}{d_a}\,E_a=\mathcal D P^{ab}_1.}
$$
Here $P^{a\bar a}_1=E_a/d_a$ 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 $E_a^2=d_aE_a$. Consequently $(E_a/d_a)^2=E_a/d_a$, as required of an <orthogonal projection>. Omitting $1/d_a$ would fail this check.
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