Let's check your idea:
$$P=\begin{pmatrix}1 & 0 & 0\\ 0 & \!\!-i & i\\ 0 & 1 & 1 \end{pmatrix}\;,\;\;P^*=\begin{pmatrix}1 & 0 & 0\\ 0 & i & 1\\ 0 & \!\!-i & 1 \end{pmatrix}$$
and since
$$PP^*\neq I\;\;\text{then}\;\;P^*\neq P^{-1}$$
The problem is this matrix's columns (rows) aren't orthonormal though they're orthogonal. We must apply Gram-Schmidt (I assume the usual euclidean inner product and let's write all the vector as row ones, for simplicity)):
$$u_1=(1,0,0)$$
$$w_2=(0,-i,1)-\langle\;(1,0,0),(0,-i,1)\;\rangle\,(1,0,0)=(0,-i,1)-0\cdot(1,0,0)=(0,-i,1)$$
and since
$$||w_2||=\sqrt 2\implies u_2=\frac1{\sqrt 2}(0,-i,1)$$
$$w_3:=(0,i,1)-\langle\;(0,i,1),(1,0,0)\;\rangle\,(1,0,0)-\left\langle\;(0,i,1),\frac1{\sqrt 2}(0,-i,1)\;\right\rangle\,\frac1{\sqrt 2}(0,-i,1)=$$
$$=(0,i,1)-\frac13\cdot 0=(0,i,1)$$
and since
$$||w_2||=\sqrt 2\implies u_3=\frac1{\sqrt2}(0,i,1)$$
Thus, our $\,P\,$ is (well, let's call it $\,Q\,$ to avoid confussion):
$$Q=\begin{pmatrix}1&0&0\\
0&-\frac{i}{\sqrt 2}&\frac1{\sqrt 2}\\
0&\frac i{\sqrt 2}&\frac1{\sqrt 2}\end{pmatrix}\;\implies\;Q^*=\begin{pmatrix}1&0&0\\
0&\frac{i}{\sqrt 2}&-\frac i{\sqrt 2}\\
0&\frac 1{\sqrt 2}&\frac1{\sqrt 2}\end{pmatrix}$$
It is now a simple exercise to verify that indeed $\,QQ^*=UI\,$ and, thus, $\,Q^{-1}=Q^*\,$ ...
Note: The above G-S process was almost trivial because the columns of $\,P\,$ were already orthogonal. The above stuff shows how to carry on the process in general, though in this case it was sufficient to divide each vector by its norm.
Step 1: Find the eigenvalues of $A$. In this case, they are $\{i,-i\}$.
Step 2: Find the corresponding eigenvectors (up to scalar multiplication). They are $(1,1)$ and $(1,-1)$. (In general, if an eigenvalue is repeated, find linearly independent and pairwise orthogonal eigenvectors corresponding to it, if possible—see the comment by @Omnomnomnom below.)
Step 3: Normalize the eigenvectors so that each has unit norm: $(2^{-1/2},2^{-1/2})$ and $(2^{-1/2},-2^{-1/2})$.
Step 4: Form a matrix $P$ having the eigenvectors as columns: $$P=\left(\begin{array}{cc}\dfrac{1}{\sqrt{2}}&\dfrac{1}{\sqrt{2}}\\\dfrac{1}{\sqrt{2}}&-\dfrac{1}{\sqrt{2}}\end{array}\right).$$
You can check that $P$ is unitary.
Step 5: Now let $D$ be the matrix having the eigenvalues in the diagonal: $$D=\left(\begin{array}{cc}i&0\\0&-i\end{array}\right).$$
Step 6: Check that $A=PDP^{-1}$.
Best Answer
You have done everything correct except for the small mistake in the last step. While normalising the vector, you have multiplied by the norm instead of dividing it and hence you are getting different values. For the first column, instead of $\frac{\sqrt{14}}{2}$ the term should be $\frac{2}{\sqrt{14}}$. And similarly for the other column. That should fix the error.