Suppose \begin{align} A=\begin{pmatrix} \frac{1}{\sqrt{n}}&\frac{1}{\sqrt{n}}&\frac{1}{\sqrt{n}}&\cdots&\frac{1}{\sqrt{n}} \\&&P \end{pmatrix} \end{align}
is a real orthogonal matrix of size $n\times n$, where the matrix $P$ is of size $(n-1)\times n$.
I have to show that $$P^\top P=I-\frac{1}{n}\mathbf1\mathbf1^{\top}\tag{1}$$
where $\mathbf1$ is an $n$ component column vector of all ones.
The choice of $P$ is certainly not unique. The most obvious choice to me is that $P$ for which $A$ is a Helmert matrix:
\begin{align} P=\begin{pmatrix} \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0&0&\cdots&0&0 \\\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{6}}&-\frac{2}{\sqrt{6}}&0&\cdots&0&0 \\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots \\\frac{1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n(n-1)}}&\cdots&\frac{1}{\sqrt{n(n-1)}}&\frac{-(n-1)}{\sqrt{n(n-1)}} \end{pmatrix} \end{align}
I could now verify that $(1)$ holds true for this $P$ but that does not prove anything.
How do I find a general form of the matrix $P$ so that $A$ is an orthogonal matrix?
Or is it possible to prove $(1)$ without explicitly finding $P$? Any hint would be great.
It is certainly possible to prove your result without finding $P$. One way to do this efficiently is to use block-matrix multiplication.
In your case, I would write $$ A = \pmatrix{\frac 1{\sqrt n} 1^T\\ P} $$ The matrices $A$ and $A^T$ are partitioned conformally. As such, we can compute $$ A^TA = \pmatrix{\frac 1{\sqrt{n}}1 & P^T}\pmatrix{\frac 1{\sqrt n} 1^T\\ P} = \frac 1n 11^T + P^TP $$ Now, $A$ is orthogonal if and only if $A^TA = I$.