My attempt:
We will use the van Kampem theorem here.
So, we take the double torus in it's "original" form, namely an octagon with directed edges ($a b a^{-1} b^{-1} c d c^{-1} d^{-1}$).
What we do, is to remove a point inside of that octagon.
So now, we want to apply the van Kampem theorem here.
We define as $U_1$ an area within that octagon (which doesn't touch the edges), and $U_2$ is an area with covers the edges of that octagon.
So here, we will have:
$$\pi_1 (U_1) \cong \pi_1 (S^1) = \langle \alpha \ | \ \alpha \alpha^{-1} = e \rangle$$
Because the edges are directed, it's simple to find out $\pi_1 (U_2)$
$$\pi_1 (U_2) \cong \pi_1 (\text{boundary of octagon with directed edges}) = \langle a,b,c,d \ | \ a b a^{-1} b^{-1} c d c^{-1} d^{-1} = e \rangle $$
This gives us:
$$\pi_1 (P \backslash \{ x \}) = \langle \alpha, a, b ,c ,d \ | \ \alpha \alpha^{-1} = e, a b a^{-1} b^{-1} c d c^{-1} d^{-1} = e \rangle$$
Did I miss something here? Or is this the correct answer?
$$\pi_1 (U_1) \cong \pi_1 (S^1) = \mathbb{Z} = \langle \alpha \rangle$$
$$\pi_1 (U_2) \cong \pi_1 (S^1 \vee S^1 \vee S^1 \vee S^1) = \mathbb{Z} * \mathbb{Z} * \mathbb{Z} * \mathbb{Z} = \langle a,b,c,d \rangle$$
$$\pi_1 (U_1 \cap U_2) \cong \pi_1 (S^1) = \mathbb{Z} = \langle \beta \rangle$$
$$\forall \ \beta \in \pi_1 (U_1 \cap U_2) \quad f(\beta) = g(\beta)$$
$f$ is a homomorphism which projects $\beta$ onto the generator of $\pi_1 (U_1)$. $g(\beta$) will give us an arbitrary combination of $a, b, c, d$, so we get:
$$\alpha = \text{combination of }a, b, c, d$$
So there is no relation here.
Thus, we are left with the elements of $U_2$, namely $a, b, c, d$, only (not $\alpha$ because we showed that it's just a combination of $a,b,c,d$). We have no relation here. So we have:
$$\pi_1 (P \backslash \{ x \} ) = \langle a,b,c,d \rangle = \mathbb{Z} * \mathbb{Z} * \mathbb{Z} * \mathbb{Z}$$