How to Perform Segue With Delay

You can achieve that by calling:

let delayInSeconds = 0.8
let segueIdentifier = "your segue id"
DispatchQueue.main.asyncAfter(deadline: .now() + delayInSeconds) { [unowned self] in
    self.performSegue(withIdentifier: segueIdentifier, sender: nil)
}

I think there is a better approach which lets the controller itself to handle dispatching issues. You can achieve it like this:
first create a method like this which later you will use its selector

- (void)showAnotherViewController{  
    [self performSegueWithIdentifier:@"yourSegueToAnotherViewController" sender:self];
}  

Then when you want to show another view controller after delay use this line of code inside your current view controller:

[self performSelector:@selector(showAnotherViewController) withObject:nil afterDelay:yourDelayInSeconds];

You can use GCD's dispatch_after to execute your segue code 2 seconds after the view appears, e.x:

- (void)viewDidAppear:(BOOL)animated 
{
    [super viewDidAppear:animated];

    double delayInSeconds = 2.0;
    dispatch_time_t popTime = dispatch_time(DISPATCH_TIME_NOW, (int64_t)(delayInSeconds * NSEC_PER_SEC));
    dispatch_after(popTime, dispatch_get_main_queue(), ^(void){
        [self performSegueWithIdentifier:@"splashScreenSegue" sender:self];
    });
}

Additionally, please make sure that you remember to call the super implementation when overriding UIViewController's life cycle methods.