Showing $P$ is a linear operator on $\ell^2$ Lets say I have defined the operator
$$P:\ell^2 \to \ell^2$$
by
$$(Pa_n)=\bigg(\frac{1}{2} \bigg)^n a_n,$$
for $a=(a_n) \in \ell^2$.

How do I show that $P$ is a linear operator on $\ell^2$?

Do we just show that $P(x+y) =P(x)+P(y)$
and $P(\alpha x)=\alpha Px$ ?

How do I show that $P$ is a bounded linear operator on $\ell^2$?
How do I determine $\|P\|$ which is the operator norm of $P$?


Yes this is how you show that $P$ is linear. 
To show it is bounded, let $(a_n)$ be a sequence in $\ell^2$. Then,
$$\|P(a_n)\|^2=\sum_{n=1}^\infty\frac{a_n^2}{2^{2n}}\le\frac{1}{4}\sum_{n=1}^\infty a_n^2=\frac{1}{4}\|(a_n)\|^2,$$
so $P$ is a bounded operator of bound at most $\sqrt{1/4}=1/2$. 
To show that $\|P\|=1/2$ consider the sequences
$$a_n=\frac{1}{2^{mn}}$$
for $m\in\Bbb N$. Then,
$$\|P(a_n)\|^2=\sum_{n=1}^\infty\frac{1}{2^{(2m+2)n}}=\frac{1}{2^{2 m+2}-1},\quad\|a_n\|^2=\sum_{n=1}^\infty \frac{1}{2^{2mn}}=\frac{1}{2^{2 m}-1},$$
so
$$\frac{\|P(a_n)\|^2}{\|a_n\|^2}=\frac{2^{2 m}-1}{2^{2 m+2}-1}=\frac{1-2^{-2m}}{4-2^{-2m}}$$
and this goes to $1/4$ as $m\to\infty$. Thus, $\|P\|^2=1/4$ and hence $\|P\|=1/2$.
