$\renewcommand{\O}{{\rm O}}
\newcommand{\GO}{{\rm GO}}
\newcommand{\PO}{{\rm PO}}
\newcommand{\PGO}{{\rm PGO}}
\newcommand{\GL}{{\rm GL}}
\newcommand{\PGL}{{\rm PGL}}
\newcommand{\SL}{{\rm SL}}
\newcommand{\SO}{{\rm SO}}
\newcommand{\Id}{{\rm Id}}
$Notation.
$K$ is an algebraically closed field;
$V$ a finite-dimensional vector space over $K$;
$f\colon V \times V\to K$ a non-degenerate symmetric bilinear form.
We write
\begin{align*}
\O(V,f)&=\{g\in\GL(V)\ |\ g^*f=f\},\\
\GO(V,f)&=\{g\in\GL(V)\ |\ g^*f =\lambda f \ \ \text{for some}\ \lambda\in K^\times\}.
\end{align*}
Here
$$(g^*f)(v_1,v_2)=f (g\cdot v_1,g\cdot v_2)\ \ \text{for}\ v_1,v_2\in V.$$
We set
\begin{align*}
\PO(V,f)&=\O(V,f)/\{\pm1\},\\
\PGO(V)&=\GO(V,F)/K^\times \subset \PGL(V).
\end{align*}
Proposition. $\PO(V,f)\cong\PGO(V,f)$.
Proof. We have the inclusion homomorphism
$$ i\colon \O(V,f)\hookrightarrow \GO(V,f)$$
which induces a homomorphism
$$i_*\colon \PO(V,f)\to \PGO(V,f),\quad\ g\cdot\{\pm1\}\mapsto g\cdot K^\times.$$
We show that $i_*$ is an isomorphism.
Let $g\cdot\{\pm1\}\in\ker i_*\,$. Then
$$g\in(K^\times\cdot\Id_V)\cap\O(V,f)=\{\pm1\}=1\cdot \{\pm1\}.$$
This $i_*$ is injective.
Let $g\cdot K^\times\in\PGO(V,f)$.
Then $ g^*f =\lambda f$ for some $\lambda\in K^\times$.
We set $g'=g/\sqrt{\lambda}$ (recall that $K$ is algebraically closed).
Then $(g')^*f=f$, whence $g'\in\O(V,f)$.
Thus
$$g\in \O(V,f)\cdot K^\times,$$
whence $i_*$ is surjective, which completes the proof of the proposition.
Concerning maximality, it is well-known that $\SO(V,f)$ is maximal in $\SL(V)$;
see, for instance YCor's answer to $\operatorname{SO}(n)$ is an (abstractly) maximal subgroup of $\operatorname{SL}(n)$
in Math.StackExchange.com in the case $K=\Bbb R$.
As above, one can deduce the maximality of $\PGO(V,f)$ in $\PGL(V)$ by standard methods.
If you cannot do that yourself, ask in MathStackExchange.com;
your question seems to be more suitable for that forum rather than for MathOverflow.
