For example in the ring $R=k[x,y]$, the ideal $I=(x,y^2)$ is $P$-primary, where $P = (x, y)$, but is not a power of $P$, so the answer to your question is negative.
We’re supposing that $I$ is an ideal in $F[[t]]$. For each $p(t)=\sum_{k\ge 0}a_kt^k\in I$, let $$a_p=\min\{k:a_k\ne 0\}\;,$$ so that
$$p(t)=a_{a_p}t^{a_p}+a_{a_p+1}t^{a_p+1}+\ldots=t^{a_p}\left(\underbrace{a_{a_p}+a_{a_p+1}t+a_{a_p+2}t^2+\ldots}_{q(t)\in F[[t]]}\right)\;.\tag{1}$$
Among all elements of $I$, choose $p\in I$ so that $a_p$ is as small as possible, and let $a=a_p$. $(1)$ shows that there is a $q(t)\in F[[t]]$ with a non-zero constant term such that $p(t)=t^aq(t)$, and the claim is that $I=(t^a)$.
Since $q(t)$ has a non-zero constant term, $q(t)$ is a unit in $F[[t]]$, and therefore $t^a\in I$; clearly this implies that $(t^a)\subseteq I$.
Now suppose that $r(t)\in I$, say $r(t)=\sum_{k\ge 0}b_kt^k$. Recall that $p$ was chosen so that $a_p$, the exponent on the first non-zero term of $p(t)$ was as small as possible for any member of $I$. That means that $a_r\ge a_p=a$. But $a_r$ is the exponent on the first non-zero term of $r(t)$, so $b_k=0$ for all $k<a_r$. And since $a\le a_r$, clearly $b_k=0$ for all $k<a$. But that means that every non-zero term of $r(t)$ has an exponent of $a$ or more, which means that we can factor out $t^a$: there is some $s(t)\in F[[t]]$ such that $r(t)=t^as(t)$. Of course this means that $r(t)\in(t^a)$, so we’ve now shown that $I\subseteq(t^a)$.
Putting the two pieces together, we get $I=(t^a)$. Thus, every ideal of $F[[t]]$ is of this form for some $a$.
Best Answer
Hint: consider the subset $\tilde{\mathbb{p}}$ of $A[[x]]$ consisting of power series $$ \sum_{n\ge 0}a_n x^n $$ where $a_n\in\mathbb{p}$. Then:
$\tilde{\mathbb{p}}$ is an ideal of $A[[x]]$
$\tilde{\mathbb{p}}$ is a prime ideal of $A[[x]]$
The contraction of $\tilde{\mathbb{p}}$ is …
You can consider the map $e\colon A[[x]]\to A$, where $e\left(\sum_{n\ge0}a_nx^n\right)=a_0$. This is a ring homomorphism.