In problems like this it can help a lot to reduce the 'size' of the problem, i.e. consider the question: What if the soccer player does not take 100 penalty kicks, but instead just 3 or 4? Let's calculate the probability distribution of the number of goals in these these scenarios explicitly:
Say the player has scored one and missed one; and is now onto kick number three. We know that they will score this kick with probability $\frac12$, thus letting $G_3$ be the amount of goals scored after $3$ kicks we see that $$\mathbb P[G_3=1]=\frac12 \hspace{1cm} \mathbb P[G_3=2]=\frac12.$$ Now let's move on to the case of the $4$-th penalty kick. Again, let $G_4$ be the number of goals scored after $4$ kicks. The probability of the player scoring on the $4$-th kick will depend on whether or not they scored the previous kick, thus a smart idea here would be to condition on the previous result $G_3$. Doing this, we obtain $$\begin{align*}\mathbb P[G_4=1]&=\mathbb P[G_4=1|G_3=1]\mathbb P[G_3=1]+\mathbb P[G_4=1|G_3=2]\mathbb P[G_3=2] \\ &= \frac23 \cdot\frac12 + 0\cdot \frac12\end{align*}$$ and similarly
$$\begin{align*}\mathbb P[G_4=2]&=\mathbb P[G_4=2|G_3=1]\mathbb P[G_3=1]+\mathbb P[G_4=2|G_3=2]\mathbb P[G_3=2] \\ &= \frac13 \cdot\frac12 + \frac13\cdot \frac12 = \frac13.\end{align*}$$ As a consequence we also get $\mathbb P[G_4=3]=\frac13$. Therefore the distribution of the number of goals kicked after $3$ penalties is uniform over $\{1,2\}$; and the distribution after $4$ penalties is uniform over $\{1,2,3\}$.
Now: Do you see the pattern? Can you prove this pattern? \
Hint:
You can prove the general pattern via Induction, i.e. compute the distribution of $G_{n+1}$ by conditioning on the result of $G_n$, just as we did above.
