I know that the same question has been asked, but I was hoping make sure that my approach was correct, and also was wondering how to proceed given my progress:
Since the projective plane is given by $aa$, and the torus is given by $bcb^{-1}c^{-1}$, it follows that $P^2\#T=(aa)(bcb^{-1}c^{-1})$. We can think of this space as a union $U\cup V$, where $U$ is the punctured torus, $V$ is the punctured protective map. The loops of $U,V$ around the punctured points are where $U,V$ are wedged together. Then by the Seifert-van Kampen theorem, the fundamental group of $P^2\#T$ is equal to $\pi_1(U,x_0)\underset{\pi_1(d,x_0)}{*}\pi_1(V,x_0)=\pi_1(U,x_0)*\pi_1(V,x_0)/\mathbb{Z}$.
cartoon of the labeling scheme and obtained surface
Now, I am wondering how to get to the actual presentation. I want to say that $\pi_1(U)=\pi_1(T)$ and $\pi_1(V)=\pi_1(P^2)$ -- is there any reason for that not to be true?
Okay so I got some help thinking this one through, so I will answer my own question.
First, we note that this puncture in the torus and the projective map makes $U$ and $V$ free groups, so now we have $\pi_1(U,x_0)=<b,c>$ and $π_1(V,x_0)=<a>$. Now we want to know what $\pi_1(d,x_0)$ is. Well, from the labeling scheme, you see that you need the loop d to be equal to both going once around the labeling scheme of $U$, which is $bcb^{-1}c^{-1}$, and once around the labeling scheme of $V$, which is $aa$, so the corresponding relation is $bcb^{-1}c^{-1}=aa$ or $bcb^{-1}c^{-1}a^{-1}a^{-1}=1$. So $\pi_1(X,x_0)=<a,b,c\mid bcb^{-1}c^{-1}a^{-1}a^{-1}>$, and we are done!