Let 𝑎 and 𝑏 be positive real numbers. Suppose $$\overrightarrow{PQ} = a\hat{i} + b\hat{j}$$ and $$\overrightarrow{PS} = a\hat{i} - b\hat{j} $$ are adjacent sides of a parallelogram 𝑃𝑄𝑅𝑆. Let $$\overrightarrow{u}$$ and \overrightarrow{v} be the projection vectors of $$\overrightarrow{u} = \hat{i} + \hat{j}$$ along $$\overrightarrow{PQ}$$ and $$\overrightarrow{PS}$$, respectively. If $$\mid \overrightarrow{u}\mid + \mid \overrightarrow{v} \mid = \mid \overrightarrow{w} \mid$$ and if the area of the parallelogram 𝑃𝑄𝑅𝑆 is 8, then which of the following statements is/are TRUE?

Adjacent sides of the parallelogram are
$$\overrightarrow{PQ}=a\hat i+b\hat j,\qquad \overrightarrow{PS}=a\hat i-b\hat j$$
with $$a\gt 0,\; b\gt 0$$.

The area of a parallelogram formed by two vectors is the magnitude of their cross-product in 3-D:
$$\text{Area}=|\overrightarrow{PQ}\times\overrightarrow{PS}|.$$

In the $$xy$$-plane we may treat $$\overrightarrow{PQ}=(a,b,0),\; \overrightarrow{PS}=(a,-b,0).$$
$$\overrightarrow{PQ}\times\overrightarrow{PS} =(a,b,0)\times(a,-b,0) =(0,0,-2ab).$$
Hence $$\text{Area}=|{-2ab}|=2ab.$$
Given area $$=8$$,
$$2ab=8\;\Longrightarrow\;ab=4.\qquad -(1)$$

The vector to be projected is taken as
$$\overrightarrow{w}= \hat i+\hat j=(1,1).$$

Projection of a vector $$\overrightarrow{w}$$ on a vector $$\overrightarrow{A}$$ has magnitude
$$|\overrightarrow{w}|\,\cos\theta =\frac{|\overrightarrow{w}\!\cdot\!\overrightarrow{A}|}{|\overrightarrow{A}|} =\frac{|\,\overrightarrow{w}\cdot\overrightarrow{A}\,|}{|\overrightarrow{A}|}.$$

The required projections are
$$|\overrightarrow{u}|=\frac{|\,\overrightarrow{w}\cdot\overrightarrow{PQ}\,|} {|\overrightarrow{PQ}|},\qquad |\overrightarrow{v}|=\frac{|\,\overrightarrow{w}\cdot\overrightarrow{PS}\,|} {|\overrightarrow{PS}|}.$$

Because $$|\overrightarrow{PQ}|=|\overrightarrow{PS}|= \sqrt{a^{2}+b^{2}},$$ we get
$$|\overrightarrow{u}|=\frac{|a+b|}{\sqrt{a^{2}+b^{2}}},\qquad |\overrightarrow{v}|=\frac{|a-b|}{\sqrt{a^{2}+b^{2}}}.$$

The condition of the problem is
$$|\overrightarrow{u}|+|\overrightarrow{v}|=|\overrightarrow{w}|=\sqrt2.$$
Therefore
$$\frac{|a+b|+|a-b|}{\sqrt{a^{2}+b^{2}}}=\sqrt2.\qquad -(2)$$

Then $$|a+b|=a+b,\;|a-b|=a-b,$$ so the numerator of (2) becomes $$2a.$$ Equation (2) gives

$$\frac{2a}{\sqrt{a^{2}+b^{2}}}=\sqrt2 \;\Longrightarrow\; \frac{4a^{2}}{a^{2}+b^{2}}=2 \;\Longrightarrow\; 4a^{2}=2a^{2}+2b^{2} \;\Longrightarrow\; a^{2}=b^{2}.$$

With $$a,b\gt 0$$ we get $$a=b.\qquad -(3)$$

Here $$|a+b|=a+b,\;|a-b|=b-a,$$ so the numerator of (2) is $$2b.$$ Exactly the same algebra applied to (2) now yields $$b^{2}=a^{2}$$ and again $$a=b.$$

Thus both cases lead uniquely to
$$a=b.$$ Using (1), $$ab=4\;\Longrightarrow\;a^{2}=4\;\Longrightarrow\; a=b=2.\qquad -(4)$$

Now each statement can be checked:

Option A $$a+b=2+2=4$$  ✔️ True.

Option B $$a-b=2-2=0\ne2$$  ❌ False.

Option C Diagonal $$\overrightarrow{PR} =\overrightarrow{PQ}+\overrightarrow{PS} =(a+a,\;b+(-b))=(2a,0).$$
Its length is $$|\,\overrightarrow{PR}\,|=|2a|=2a=4.$$ ✔️ True.

Option D An internal angle bisector of $$\overrightarrow{PQ}$$ and $$\overrightarrow{PS}$$ is in the direction $$\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}+ \frac{\overrightarrow{PS}}{|\overrightarrow{PS}|} =(1/\sqrt2,1/\sqrt2)+(1/\sqrt2,-1/\sqrt2) =( \sqrt2,0),$$ i.e. along the positive $$x$$-axis. $$\overrightarrow{w}=(1,1)$$ is parallel to $$\overrightarrow{PQ}$$, not to this bisector. ❌ False.

Therefore the correct statements are:
Option A and Option C.