1. Given vectors

The adjacent sides of parallelogram $PQRS$ are $\overrightarrow{PQ}=a\hat i+b\hat j,$ $\overrightarrow{PS}=a\hat i-b\hat j,$ with $a,b>0$.

Also, $\vec w=\hat i+\hat j.$

Let $\vec u$ and $\vec v$ be the projection vectors of $\vec w$ along $\overrightarrow{PQ}$ and $\overrightarrow{PS}$ respectively.

2. Area of the parallelogram

Area is the magnitude of the 2D cross product: $\text{Area}=\left|\begin{vmatrix} a & b \\ a & -b \end{vmatrix}\right|=| -ab-ab|=2ab.$

Given area $=8$, so $2ab=8 \implies ab=4.$

3. Projection lengths of $\vec w$ on $\overrightarrow{PQ}$ and $\overrightarrow{PS}$

For a vector $\vec x$ projected on direction $\vec d$, the magnitude of projection is $\frac{|\vec x\cdot \vec d|}{|\vec d|}.$

Since projection vector magnitude equals scalar projection magnitude here,

Projection on $\overrightarrow{PQ}$

$\vec w\cdot \overrightarrow{PQ}=(1,1)\cdot(a,b)=a+b.$ Also, $|\overrightarrow{PQ}|=\sqrt{a^2+b^2}.$ Thus $|\vec u|=\frac{a+b}{\sqrt{a^2+b^2}}.$

Projection on $\overrightarrow{PS}$

$\vec w\cdot \overrightarrow{PS}=(1,1)\cdot(a,-b)=a-b.$ So the projection vector magnitude is $|\vec v|=\frac{|a-b|}{\sqrt{a^2+b^2}}.$

Given $|\vec u|+|\vec v|=|\vec w|=\sqrt2.$ Hence, $\frac{a+b+|a-b|}{\sqrt{a^2+b^2}}=\sqrt2.$

Now consider cases.

4. Solve the condition

Case 1: $a\ge b$

Then $|a-b|=a-b$, so $\frac{a+b+a-b}{\sqrt{a^2+b^2}}=\sqrt2$ $\frac{2a}{\sqrt{a^2+b^2}}=\sqrt2.$ Squaring, $\frac{4a^2}{a^2+b^2}=2$ $4a^2=2a^2+2b^2$ $a^2=b^2.$ Since $a,b>0$, this gives $a=b$, which is a subcase of $a\ge b$. Then with $ab=4$, $a=b=2.$

Case 2: $a<b$

Then $|a-b|=b-a$, so $\frac{a+b+b-a}{\sqrt{a^2+b^2}}=\sqrt2$ $\frac{2b}{\sqrt{a^2+b^2}}=\sqrt2.$ Similarly, $b^2=a^2 \implies a=b,$ contradicting $a<b$ unless again $a=b$.

So the only possibility is $a=b=2.$

5. Check each option

Option A: $a+b=4$

Since $a=b=2$, $a+b=2+2=4.$ So A is true.

Option B: $a-b=2$

$a-b=2-2=0\ne 2.$ So B is false.

Option C: Length of diagonal $PR$ is 4

For a parallelogram, diagonal $\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{PS}.$ So $\overrightarrow{PR}=(a\hat i+b\hat j)+(a\hat i-b\hat j)=2a\hat i.$ Hence $|PR|=2a=2\cdot 2=4.$ So C is true.

Option D: $\vec w$ is an angle bisector of the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PS}$

With $a=b=2$, $\overrightarrow{PQ}=2\hat i+2\hat j=2(\hat i+\hat j)=2\vec w.$ So $\vec w$ is actually along $\overrightarrow{PQ}$, not the bisector between $\overrightarrow{PQ}$ and $\overrightarrow{PS}$.

Also, $\overrightarrow{PS}=2\hat i-2\hat j.$ The angle between $\overrightarrow{PQ}$ and $\overrightarrow{PS}$ is $90^\circ$, so the internal bisector would be along the $x$-axis, i.e. along $\hat i$, not along $\hat i+\hat j$.

Therefore D is false.

6. Final answer

The true statements are: $\boxed{\text{A, C}}$

This matches the stored correct answer.