= Solution
Interpret the paired $A,B,C$ disks as the three <one-handles> of a genus-three <handlebody>. Attach <two-handles> on the two displayed $\beta$ curves. Orient the three disk-crossing generators as $a,b,c$. Reading the signed crossings, starting at the upper-left portion of each attaching curve and changing the start point when needed, gives
$$
r_1=a b c a^{-1}c^{-1}b^{-1},\qquad r_2=a b c b^{-1}a^{-1}c^{-1}.
$$
The first says $abc=bca$ and the second $abc=cab$. Thus the <fundamental group> presentation is
$$
\boxed{\pi_1(Y)=\langle a,b,c\mid [a,bc]=1,\ [ab,c]=1\rangle.}
$$
Here $[u,v]=uvu^{-1}v^{-1}$. Cyclically changing the starting point, reversing an attaching curve, or changing generator orientations gives equivalent presentations. With $h=abc$, these relations make $h$ central; eliminating $c=b^{-1}a^{-1}h$ gives
$$
\pi_1(Y)\cong\langle a,b,h\mid[a,h]=[b,h]=1\rangle\cong F_2\times\mathbb Z.
$$
For the topological identification, thicken the diagram's two nested bands and identify the three paired disk mouths. The complement of those bands is the product region of a <pair of pants> with a circle; its two compressing curves are exactly $[a,bc]$ and $[ab,c]$. Equivalently, the standard cell decomposition of this product has three <one-handles> and the two commuting <two-handle> attachments shown. Thus this is a <generalized Heegaard diagram> of $P\times S^1$.
The <Hopf fibration> of $S^3$ has three disjoint regular fibers whose removal leaves $P\times S^1$. Its three fibers are the components of the <torus link> $T_{3,3}$, as is also apparent from the full three-strand twist in the later <link diagram>. This identifies $Y$ with that <link exterior>, using the diagram and product structure rather than just its <fundamental group>. In particular \b[it is a link exterior in the three-sphere].
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